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2.2. 



GENERAL PROBLEMS 



OP 



SHADES AND SHADOWS 



FORMED BOTH BY PARALLEL AND BY RADIAL RAYS ; AND 

SHOWN BOTH IN COMMON AND IN ISOMETRICAL 

projection: TOGETHER WITH THE 

THEORY OF SHADING. 



^ 



- 



S* EDWARD WARREN, C.E., 

PROF. OF DESCRIPTIVE GEOMETRY, ETC., IN THE RENSSELAER POLYTECHNIC INSTITUTE; AND AUTHOP 

OF " ELEMENTARY PLANE PROBLEMS j" " DRAFTING INSTRUMENTS, ETC. j" " ELEMENTARY 

PROJECTION DRAWING;" "ELEMENTARY LINEAR PERSPECTIVE;" 

AND " DESCRIPTIVE GEOMETRY." 



SECOND THOUSAND. 




<f/Z? a> 



NEW YORK: 

JOHN WILEY & SONS, PUBLISHERS, 

53 East Tenth Street. 

1895. 



K=» 



Entered according to Act of Congress, in the year 1867, by 

JOHX WILEY & SON. 

In the Clerk's Office of the District Court of the United States for the Southern District 

of Xew York. 

Copyright renewed 1895. 



S\\ 



^ 



^ CONTENTS 



General Table i^ 

Preface xi 

BOOK I. — Construction of Shades and Shadows 

PART I. — In common projection 



SERIES I.— By parallel rays. 



General Principles 

§ 1. — Definitions and classification of problems 

§ 2. — Graphical representation of Shades and Shadows. 



DIVISION I. — Shades and Shadows on Ruled Surfaces. 
Class!.— " " Plane " 



§ I. — Shades 

Problem I. — To determine, first whether the fine of intersection 
of two oblique planes is, or is not, a line of shade ; and, 
second, to determine, by inspection, the compound line of 
shade on a given plane sided body 8 

Problem II. — To determine the edges of shade on a pyramid, 
whose axis is oblique to both planes of projection. — Two 
methods 9 

§ IL— Shadows 11 

Problem III. — To find the shadow of a straight line on the hori- 
zontal plane of projection 11 

Problem IV. — To construct the shadow of a square abacus upon a 
square pillar, and of both, on the horizontal plane of projec- 
tion 12 

Problem V. — Having the projections of a ray of light, upon two 
vertical planes at right angles to each other ; to find its projec- 
tions on those planes, after revolving it 90° about a vertical 
axis 13 

Problem VI. — To find the projections of the shadows on the parts 
of a timber framing, shown in two elevations, on two vertical 
planes at right angles to each other 14 



IT CONTEXTS. 

PA&l 

Pboblem TIL — To construct the shadow of a circle on the vertical 

plane of projection. 16 

Pboblem Till. — To construct the shadow of a straight Hie on a 

plane whose traces are parallel to the ground line IP 

Pboblem LX — To find the shadow of a chimney, situated on the 

end portion of a hipped roofj upon the end and side roofs . . 19 
Problem X — To find the shadow of a shelf and brackets upon a 

vertical plane 22 



CLASS II. — Shaiies axd ;iir:w; o>~ 5i>-g-lb Ccsved Ststaces. rs* 

Gekeeal 23 

SECTION I. — 0~. Li:-:~,:r>j.'b',i S:^f,i C :..■'': -:d S:..-fo.c~:.s 23 

§ I.— Shades 23 

Pboblem XI. — To construct the elements of shade on any cylinder 

or cone 24 

MrsL — On a horizontal right semi-cylinder 24 

Second. — On any oblique cylinder. . 24 

Third. — On a cylinder, whose axis is parallel to the 

horizontal plane only 24 

Fourth, — On a cone 25 

Fifth. — On two intersecting cylinders 26 

§ II— Shadows 2T 

Pboblem XII. — To construct the shadow of the projecting head of 

a cylinder upon the cylinder 21 

Problem XIII. — To construct the shadow of the edges of shade 

of a cross upon a cylinder, both bodies being seen obliquely . 28 
I s : 3LEM XIV. — To construct the shadow cast by the upper base 

of a hollow oblique cylinder upon its interior. 30 

Pboblem XT. — To construct the shadow of the upper base of a 
vertical right cone, upon the lower nappe of the same 

cone 32 

Pboblem XTX — To construct the shadow cast by the vertical cylin- 
der (Prob. XI., Fifth) upon the horizontal cylinder, and by 

the horizontal cylinder upon the vertical cylinder 34 

1°.— Of the element of shade of the horizontal 

cylinder on the vertical cylinder 35 

2°. — Of the upper base of the vertical cylinder 

upon the horizontal cylinder 35 

3°. — Of the element of shade of the vertical cylin- 
der upon the horizontal cylinder 36 

Pboblzm XVII — To find the axes of an elliptical shadow, two of 

whose conjugate diameters are known 37 

Pboblem XV III. — Having a vertical semi-cyhnder with its meridian 
plane parallel to the vertical plane of projection, it is required 



CONTENTS. V 

PAGH 

to find the shadow cast on its base and visible interior, by 
its edge of shade, and by a vertical semicircle, described on 

the diameter of its upper base 3? 

1°. — The shadow of the vertical edge 39 

2°. — " " front semicircle 39 

3°. — " on the upper base circle 39 

4°. — " by auxiliary spheres 39 



SECTION II. — Shades and Shadows on Warped Surfaces 41 

§ I.— Shades 41 

Problem XIX. — To construct the curve of shade on a hyperbolic 

paraboloid 42 

Problem XX. — To find the curve of shade on the common oblique 
helicoid, in the practical case of the threads of a triangular- 
threaded screw 45 

1°. — Description of the screw - . . 45 

2°. — To construct any particular element. (Two 

methods.) 4G 

Points of shade on assumed helices 46 

First. — By planes of given declivity. 47 

1°. — On an outer helix. 
2°._ " inner " 

Second. — By helical translation 48 

Points of shade on assumed elements. Discussion 52 

Problem XXI. — To construct points of the curve of shade of a 

conoid. Discussion 55 

§ II. — Shadows 58 

Problem XXII. — To find the several shadows cast by a triangular- 
threaded screw upon itself 60 

First— Of the nut 60 

1st. — Of any unknown point of an edge. ........ 60 

2d. — Of any assumed point of an edge 62 

Second. — Of the curves of shade 63 

1st. — On an assumed element 63 

2d. — On an assumed helix 64 

3d. — On another branch of the curve of shade. ... 64 

Third— Of the outer helix 64 

1st. — On an assumed element 64 

2d. — On any helix 65 

3d. — On particular elements „ . 65 

Fourth. — The shadow on the horizontal plane 66 

1st. — The determination of the fines casting this 

shadow 66 

2d. — The Shadow of any point 66 

Problem XXIII. — To determine, in general, the shadows of a ver- 
tical right conoid 66 



VI CONTENTS. 



PAGB 

DIVISION II.— Shades and Shadows on Double Curved 

Surfaces 68 

§ I.— Shades , 68 

Problem XXIV. — To find the curve of shade of the sphere, using 

only one projection of the sphere 69 

Problem XXV. — To find the curve of -shade on an ellipsoid of re- 
volution 71 

Problem XXYI. — To find the projections of the curve of shade on 

an ellipsoid of three unequal axes 72 

Problem XXYII. — To construct the curve cf shade upon a torus. 75 

1°. — Four points on the visible boundaries 75 

2°. — The highest and lowest points 75 

3°. — Intermediate points 76 

Problem XXY1II. — To find the curve of shade on a piedouche. . . 76 

1 °. — Preliminary points of shade 77 

2°. — Intermediate points, by tangent cones 78 

3°. — " by tangent spheres 79 

§ II. — Shadows 81 

Problem XXIX. — To construct the shadow of the front circle of 

a niche upon its spherical part 82 

1°.— First Solution. General Method 82 

2°. — Second Solution. Special Method 83 

3°. — Point where the shadow leaves the spherical 

part 84 

4°. — Third Solution. Special Method. Discussion. 

Various constructions of e 84 

Problem XXX. — To find the shadow of the upper circle of a pie- 
douche upon its concave surface 88 

1°. — The highest and lowest points of shadow. ... 88 

2°. — Any intermediate points. Discussion 88 

§ III — General Problem in Review of Shades and Shadows. 

determined by Parallel Rays 91 

Problem XXXI. — To find the shades and shadows of the shaft 

and base of a Eoman Doric Column 92 

Problem XXXII. — To find the shades and shadows of the capital 

and shaft of a Eoman Doric Column 95 



SERIES II. — Shades and Shadows determined by diverging rays 97 

Section I. — General Principles 97 

" II. — Problems involving diverging rays 99 

Problem XXXIII. — To find the shadow of a semi-cylindrical 

abacus, upon a vertical plane through its axis 99 



CONTENTS. V13 

PAGBi 

Problem XXXIV. — To find the curve of shade on a sphere, the 

light proceeding from an adjacent point IOC 

Problem XXXV. — Having a niche, whose base is produced, form- 
ing a full circle ; and a right cone, the centre of whose cir- 
cular base coincides with the centre of the base of the niche, 
it is required to find the shades and shadows of this system, 
when illumined by an adjacent point 101 



PART II. — Shades and Shadows in Isometrical Projection 102 

Section I. — General principles 103 

Problem XXXVI. — To find the angle made by the isometrical ray 

of light with the isometrical plane of projection 104 

Section II. — Shades and shadows on isometrical planes 105 

Problem XXXVII. — Having given a cube, with thin plates pro- 
jecting vertically and forward, in the plane of its left hand 
back face, to find the shadows of the edges of these plates 

upon the cube and its base 105 

Section III. — Shades and shadows on non-isometrical planes 106 

Problem XXXVIII. — To find the shadow of a hexagonal cupola 
on a coupled roof, one face of the cupola making equal angles 

with two adjacent walls of the house 106 

Section IV. — Shades and Shadows on single curved surfaces 10 1 ? 

Problem XXXIX.— To find the elements of shade on an inverted 
hollow right cone, the shadow on its interior, and the shadow 

of one of its elements of shade on an oblique plane 107 

Section V. — Shades and Shadows on double curved surfaces 108 

Problem XL. — To find the curve cf shade on a sphere 108 

Problem XLI. — To construct the curve of shade on a torus 109 



BOOK II. — The finished execution of shades and shadows 112 

CHAPTER I. — Theory and construction of brilliant points ; and of 

gradations of shade 112 

SECTION I. — Preliminary general principles 112 

§ 1°. — Geometrical conditions for the adequate graphical representa- 
tion of form 11? 

§ 2°. — Of the physical conditions for the visibility of bodies, and an 

adequate representation of their forms 1 IS 

§ 3°. — Of brilliant points and lines. Ill 

SECTION II. — The construction of brilliant points and lines ... 118 

§ 1. — Brilliant elements of planes 118 



Vlll CONTENTS. 

Men 

PEOBiiEM XLII. — To find the brilliant point of a plane which re- 
ceives light from a near luminous point 119 

§ 2. — Brilliant elements on developable surfaces 119 

Problem XLin. — To find the brilliant element on a cylinder, illu- 
minated by parallel rays. (Two solutions) 119 

Problem XLIY. — To find the brilliant point of a cylinder which is 

ilhiminated by diverging rays 121 

Pboblem XLV. — To find the brilliant point on a cone which is illu- 
minated from a near luminous point 122 

§ 3. — Brilliant points on warped surfaces 123 

Problem XLVT. — To find the brilliant element of a warped surface 

when illuminated by parallel rays 123 

Problem XLV1L. — To construct the brilliant point on a screw, 

when illuminated by parallel rays 124 

§ 4. — Brilliant points on double curved surfaces 125 

Problem XLV HI. — To find the brilliant point on any double-curved 

surface, when illuminated by parallel rays 125 

Problem XLTX. — To find the brilliant point on a sphere, illuniinat- 

ed by parallel rays 125 

Problem L. — To find the brilliant point on a piedouche 126 

CHAPTER EL — Of the representation of the gradations or light 

AND SHADE... 127 



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PREFACE. 



Wherefore thus pursue a shadow, some may ask, seeing the title of 
this volume, and the complexity of some of its figures. 

We will endeavor to reply. 

The study of Shades and Shadows is an application of the general 
problems of Descriptive Geometry, in connexion with a few physical 
principles ; and, of Descriptive Geometry, no one, who has occasion to 
be conversant with forms, singly or in combination, can know too 
much, either for practical purposes, or as a promoter, in its peculiar 
way, of mental power. 

In particular, having in a previous work on the General Problems of 
Descriptive Geometry, disposed of the Projections, Intersections, Tan- 
gencies, and Developments of geometrical forms in the abstract, the 
way is prepared for the extended direct application of the second and 
third of these operations, involving incidentally that of the first, to the 
concrete geometrical subject of Shades and Shadows. For a Shade is 
always bounded by a line of contact, and a Shadow by one of intersec- 
tion ; and before either can be found, the body casting, and the surface 
receiving the shadow, must be given by their projections. 

The easy logical sequence of " General Descriptive " Problems, and 
" Shades and Shadows " as a theoretical branch of applied " Descrip- 
tive," is, therefore, one point of interest attaching to the study of the 
latter subject. 

Further, there is beauty in the idea, that for any distance and direc- 
tion of the source of light, and for any form and position of the bodies 
casting and receiving shadows, the mind can know, and the hand can 
execute those shadows truly. 

The utility, however, of delineated shades and shadows, in rendering 
working drawings at once not only more beautiful, but more intelligible, 
because more nearly conformed to reality, is the chief ground of inte- 



x:: pez?^:^ 

rest in the study of them. The student, therefore, who lias experienced 
any degree of pleasure in the pursuit of the general problems ot 
u Descriptive, w mar, without probable disappointment, promise himself 
added enjoyment in the study here introduced to him. 

Once more : The many interesting theoretical and practical par:: : n- 
lars in which the science of optics is found related to the special study 
of shades and shadows, lend interest to the latter, as affording a field 
for observation and reflection to be enjoyed in common by the artist, 
the physicist, and the geometrical draftsman. Indeed, it is hoped that 
Book EL of this volume, especially Chapter II., on the execution of 
shades and shadows, will be found worthy of the attention of arrises, 

A word now as to the plan of this work. 

'Treat care has been bestowed on the statements of general principles, 
which, in any effective study of the subject, must be mastered as fully 

Again: with his mind properly guarded against perplexity, by due 
comprehension of a simple, and manifestly rational, classification of 
the problems of a proposed course, the student will gain little effective 
assistance from the study of a protracted series of similar problems. 

One good characteristic problem, under each of the heads poin'.r i 
out in the preceding general table, would therefore be, at least in 
theory, sufficient. 

He student having mastered these representative problems, would 
then readily refer any new problem, met by him, to its appropriate 
group ; and would then solve it by an intelligent use of the generai 
principles applicable to all problems of that group. 

But it is usually difficult for the beginner to dispossess himself 
entirely of the idea that problems, which differ considerably in out- 
ward form, do not differ likewise in essential principle, and method of 
solution. Besides, there are often indirect and special methods oi 
solution which really need separate exemplification ; hence, a liberal 
number of special problems, some of them of quite a practical cast, 
have been scattered through this work. 

With this modification of what was at first pointed out as a theoreti- 
cally sufficient course, it is hoped that the present effort will be accept- 
able to professors and artists, and to classes in scientific collegiate 
courses, and in the Polytechnic Schools ; especially to those who have 
already become familiar with the rudiments of Shades and Shadow ; 



PREFACE. Xlii 

presented in the author's "Elementary Projection Drawing," or in 
similar works. 

As to methods of study, none is better, after reading the text till it 
is fully understood, than repeated rehearsals of the problems, and prin- 
ciples, by the student, with the book closed ; the figures having been 
made upon a slate or portable blackboard, such as every student should 
have, and in any form which will truly represent the essential opera- 
tions of the problems, without regard to the particular forms of the 
figures in the book. 

The recitations may properly consist of mingled interrogation upon 
the principles, and explanations of the problems ; the latter, both from 
the figures in the book, and from blackboard constructions ; also from 
the finished constructions on the student's plates, when time enough is 
devoted in the class-room to their construction, to make due interroga- 
tion upon them possible. It will, however, probably appear on trial 
that no test equals the regular class recitation, in that thoroughness 
which is due to the student in his training. 

It is a peculiar pleasure to add, that this expensive volume now 
appears, through the every way timely and pleasant kindness of Students 
of the Institute, to the Author, and, as he truly hopes, no less to them- 
selves, in making up a liberal subscription in aid of its hitherto delayed 
Duplication. 



SHADES AID SHADOWS. 



Book 1. 

CONSTRUCTION OF SHADES AND SHADOWS. 

PART I. 
IN COMMON PROJECTION. 

SERIES L 

SHADES AND SHADOWS FORMED BY PARALLEL RATS. 

GENERAL PRINCIPLES. 
§ 1. — Definitions, and Classification of Problems. 

1. Light proceeds from its source in uninterrupted straight 
lines, called rajs, except when diverted by reflection or refrac- 
tion, or arrested by opaque bodies ; as is sufficiently proved by 
the impossibility of direct vision through a bent tube. 

2. If the source of light be a point, the volume of rays pro- 
ceeding from it, and intercepted by an opaque body, will be a 
cone, having the luminous point for its vertex. If this point be 
at an infinite distance from the opaque body, the cone becomes a 
cylinder, and the rays of light will be sensibly parallel. The latter 
case is the one considered in the present series of problems, and 
is substantially the same as that of bodies illuminated by the 
Sun, for, on account of his great distance from the Earth, the 
solar rays to all points of any terrestrial object, are sensibly parallel. 

3. Let there now be given, in the following order of successive 
position, a source of light, an opaque point, and an opaque sur- 
face. The opaque point will intercept one of the rays from the 
source of light. The path of the ray, beyond the opaque point, 

l 



2 GENERAL PROBLEMS. 

will therefore be a line of darkness, and its intersection with the 
opaque surface will be a dark point. 

The line of darkness is the shadow in space of the given point ; 
the dark point on the opaque surface is the superficial shadow, 
or, simply, the shadow of the same point. 

4. The shadow of a point upon any surface is, therefore, the 
intersection of a ray (meaning the indefinite path of the ray) 
drawn through the point, with the given surface. 

5. Any plane containing a ray, is a plane of rays ; for any 
number of rays can be drawn in the plane, and parallel to the 
given ray, and these rays will, collectively, constitute a plane of 
rays. Hence a plane of rays can be drawn through a straight 
line having any position in space. For the plane containing 
this line, and a ray through any point of it, will be the required 
plane of rays. Furthermore, since two lines determine a plane, 
but one plane of rays can be passed through a given line, unless 
the line be parallel to the rays of light. 

6. A line being made up of points, its shadow will consist of the 
shadows of all its points. Rays of light being parallel, the rays 
through all the points of a straight line, will form a plane of rays 
(5), unless the line coincides with a ray. The shadow of a straight 
line is, therefore, in general, the line of intersection of a plane 
of rays passed through it, with the surface receiving the shadow. 
When the line coincides with a ray, its shadow is its own point 
of intersection with the given surface. 

When a line is parallel to any plane, its shadow on that plane 
is parallel to the line itself. 

The shadows of parallel lines, on the same plane, are parallel. 
Also the shadows of the same line on parallel planes are parallel ; 
only one of them, however, can be real. 

7. The parallel rays through all points of a curve, will form a 
cylinder of rays, unless the curve be plane, and coincides with a 
plane of rays. Hence the shadow of a curved line is, in general, 
the intersection of the cylinder of rays having the curve for its 
directrix, with the given opaque surface. When the curve is 
plane, and in a plane of rays, its shadow is the line in which 
that plane intersects the opaque surface. 

8. The shadow of a surface is a surface, unless it is a plane 
surface and coincides with a plane of rays, when its shadow will 
be a line, as in the case of a plane curve lying in a plane 
of rays. 



SHADES AND SHADOWS. 3 

9. The shadow of a solid is always a surface, which is known 
when its boundary, called its line of shadow, is found. Let the 
given body be circumscribed by parallel tangent rays. These 
rays will form a circumscribing tangent cylinder of rays. The 
curve of contact of this cylinder and the body, is evidently the 
boundary between the side of the body towards the source of 
light, and the part which is opposed to the light. The former is 
the illuminated part of the body, the latter is the shade of the 
body, and the line of contact is called the line of shade. 

10. Therefore 1st. From all the portion of space within the tan- 
gent cylinder of rays, and beyond the given body, light is exclud- 
ed, and that portion is called the shadow in space, of the body. 
2nd, The area, bounded by the intersection of the cylinder of 
rays with the surface receiving the shadow, is the shadow of the 
body on that surface. 3c?. The line of shadow is thus the shadow 
of the line of shade. 4zth. The line of shade on a body must 
therefore be known, before its line of shadow can be found. 

11. Any one point of a curve of shade upon a body, is the 
point of contact of one element of the tangent cylinder of rays ; 
that is of one ray : and the intersection of that ray, with any 
surface beyond the given body, is the shadow of that one point of 
the curve of shade. Both shades and shadows are found, in 
practice, according to this statement, i.e. by separately rinding 
single points and joining them. 

12. The constitution, and mode of construction, of the line of 
shade of a body, will depend on the nature of the surface of 
that body. The line of shade on a plane sided body, is com- 
posed of those edges, collectively, which divide its light, from its 
shaded surfaces. On a developable single curved surface, it con- 
sists of the elements of contact of tangent planes of rays. On 
a warped surface, it consists of a succession of points, one on 
each of a succession of elements, these points being the points 
of tangency of planes of rays. On a double curved surface, it 
consists strictly of the curve of contact of a circumscribing tan- 
gent cylinder of rays. 

The problems of shades may therefore be arranged in the order 
of the four kinds of surfaces as just named. 

13. Every problem of shadows, evidently reduces to rinding 
the shadow of a point on a given surface ; that is, to finding the 
intersection of a straight line (ray) with the surface. The 
method of rinding this intersection depends on the nature of the 



4 GENERAL PROBLEMS. 

given surface, hence the problems of shadows may be arranged in 
the same way as those of shades may be 

But the construction of the shadows of straight lines beins 
generally a little simpler than that of the shadows of curves, 
the surface receiving the shadow being the same in both cases, 
the former may be made introductory to the latter, in each of the 
four principal groups of problems of shadows. 

1-i. Rehearsing the principal conclusions thus far reached, we 
have the following: 

The curve of shade- of a body is the curve of contact of a tangent 
ider of rays, with that body. The curve of shadow of the 
same body, is the curve of intersection of the same cylinder with 
the surface receiving the shadow. 

Any one point of the curve of shade, is the point of contact of 
element of the cylinder, and this element is a ray of light The 
intersection of the same ray, or element, with the surface receiving 
the shadow, is the shadow of this same point in the curve of shade. 

15. Here, expanding the last article somewhat, we observe 
that the line of shade is either straight or curved, hence the sur- 
face of rays passed through it — and whose intersection with auy 
other surface is the line of shadow — will always be a plane when 
the line of shade is straight, and sinsrle curved when the line of 
shade is a curve. In the latter case the line of shade will be the 
directrix of the single curved surface of rays. "When the i 
are parallel, the single curved surface of rays will be cylindrical; 
when they are diverging, it will be conical, having the luminous 
point for its vertex. Hence, for all combinations of straight and 
curved lines of shade, with parallel rays, and rays diverging 
from a point, the curve of shadow — defined as found by the 
problems of orthographic projection — will be the intersection of a 
plane, a cylinder, or a cone, with the surface receiving the shadoic. 

16. The case, in which the source of light is a luminous point 
either at a finite or infinite distance, is the only one which needs 
to be considered. For, by way of a simple example of simui. 
taneous illumination from several points, let a sphere be exposed 
to the light from a finite luminous straight line. Each point of 
such a line would be the vertex of a separate tangent cone of 
rays, whose circle of contact with the sphere would be the circle 
of shade due to that cone. Hence, onlv so much of the sphere 
as received no light from either extremitv of the luminous line, 
would be totally in shade ; and onlv so much as received li°rht 



ri.i. 





I'M. 



SHADES AND SHADOWS. 

from every point of the line would be totally illuminated. The 
intermediate portion would be in partial shade, which would 
become more and more complete as the total shade should be 
approached. In order, therefore, to secure both simplicity and 
precision in the diagrams of shades and shadows, the source of 
light is assumed to be a point. 



§ 2. — Graphical Representation of Shades and Shadows. 

17. In passing to the graphical construction of shadows, it is 
to be observed, that, in representing them in orthographic projec- 
tion, the planes of projection, the projecting lines, and the use of 
these, are similar to the same things, and their use, as found in 
any other problems of orthographic projection. 

18. The three things given in space (3) for the production of a 
real shadow, are given in projection, in representing that shadow ; 
only, as the source of light is supposed to be indefinitely remote, 
it is only virtually represented, by the projections of the parallel 
rays proceeding from it. 

If the source of light were a near luminous point, its projec- 
tions would be given, together with rays proceeding from it in 
diverging lines. 

19. The ordinary rules of orthographic projection, concerning 
given and auxiliary, hidden and visible lines and planes, are 
observed in problems of shades and shadows. Moreover, visible 
shades or shadows are represented, either by a tint of indian ink 
or by parallel lines. Hidden shades or shadows are represented 
only by their boundaries, which are made in dotted lines ; plain, 
or fringed, as in PI. L, Fig. 5, for greater distinctness. 

20. In constructing shadows merely as a geometrical exercise, 
the main object is to indicate clearly their position. Neatness 
and distinctness are, therefore, all that is required in their exe- 
cution ; hence, to save time, they are represented in flat or uni- 
form tints. In making finished drawings, the shades and shadows 
are done in graduated tints, according to their actual appearance. 

21. In industrial applications of problems of shades and sha- 
dows, the light is usually taken to correspond with the body 
diagonal of a cube, two of whose faces coincide with the planes 
of projection. That is, the light coming from the left, forward 
and downward, makes an angle of 35° 16' with the planes of 



6 GENERAL PROBLEMS. 

projection, and its projections make angles of 45° with the 
ground line. 

In the following general problems, the light is taken indif- 
ferently in any direction, since the general methods of solution 
are not altered by so doing. 

22. Methods of solution are distinguished as General and Spe- 
cial, also as Direct and Indirect. 

General methods consist of such operations as apply equally 
well, in theory, though not always with equal practical conve- 
nience, to whole classes of problems. 

Special methods are such as are founded on the distinguishing 
peculiarities of individual 'problems, or groups of problems. They 
are therefore as numerous as those peculiarities, and hence afford 
the natural field for the exercise of ingenuity in their discovery 
and application. 

Direct methods are those which immediately and obviously 
conform to the usual definition of a shade or a shadow ; as when 
a point of shade is found as the point of contact of a given ray 
with a given surface. 

Indirect methods are those in which auxiliary magnitudes are 
employed, to avoid the laborious constructions which direct 
methods may sometimes occasion ; as when we find a point of 
shadow on some irregular surface, by noting where the easily 
found shadow upon some secant plane meets the intersection of 
that plane with the surface. 



SHADES AND SHADOWS. 



SERIES I. 

SHADES AND SHADOWS FORMED BY PARALLEL RAYS 

DIVISION I. 

SHADES AND SHADOWS ON RULED SURFACES. 

23. The line of shade on any ruled surface is obtained by find- 
ing, either by inspection, or by construction — by means of tangent 
rays or planes of rays — the edges, elements, or points which 
constitute that line of shade. 

24. A shadow cast by a body upon any ruled surface, is deter- 
mined by finding an element of that surface, and a point of the line 
of shade of the body, both of which shall be in the same plane 
of rays. A ray through the point of shade will then intersect 
the element of the ruled surface in a point of the required shadow. 

CLASS I. 

SHADES AND SHADOWS ON PLANE SURFACES. 

§ I. — Shades. 

25. The shade of an opaque plane, or plane figure, is the whole 
of one side of such plane or plane figure. The former being of 
indefinite extent, can have no line of shade. That of the latter 
is its perimeter. 

26. The line of shade of any plane-sided body, as a pyramid, 
consists of those edges, collectively, which divide its illuminated 
from its unilluminated surfaces (12). This line of shade can 
often be determined by inspection, having given the relative 
position of the body and the rays of light. 

The two following problems will show the method of finding 
lines of shade on plane-sided bodies, when they cannot be deter- 
mined by inspection. 



8 GENERAL PROBLEMS. 



Problem I. 

To determine, first, whether the line of intersection of two oblique 
planes is, or is not, a line of shade ; and, second, to determine, by 
inspection, the compound line of shade on a given plane-sided body. 

First: — Principles. — After finding the projections of the line of 
intersection of the given planes, by Prob. Y., Des. Geom., assume 
any point on that line, and draw the projections of a ray of light 
through it. If the ray, thus given, pierces either plane of pro- 
jection between the traces of the given planes on that plane, it 
shows that both planes are illuminated, both being pierced by 
rays, and hence that their intersection is not a line of shade. On 
the contrary, if the ray pierces either plane of projection outside 
of the traces of the given planes on that plane, it shows that the 
intersection of the given planes is a line of shade. 

Construction. — PI. I., Fig. 1. Let RS be the ground lin^ 
PEP' one of the given planes, PSP' the other, and P/--PV 
their line of intersection, n, n' is an assumed point of this inter- 
section, through which the ray, na—n'a', parallel to the given 
ray, r— r\ is drawn. This ray pierces the horizontal plane at 
a, a', indicating, according to the principles of this solution, that 
P/— PV is a line of shade. 

Remarks. — a. Hence the projections of the visible portion of 
the unilluminated plane, PSP', are shaded, and P/— P'e' is inked 
as a heavy line. 

b. The above construction is applicable to the case of any 
pyramid, in a simple, or oblique position, when we wish to deter- 
mine its line of shade. In such cases, unless the pyramid rests 
upon a given plane, it will first be necessary to determine the 
traces of its faces on some plane, whose intersections with the 
rays through points in the edges of those faces will also be con- 
structed. 

27. As a second illustration of this problem, see PI. IY., Fig. 
14, where PQP' and PRP' are the given oblique planes. P'a' 
and Vb are the projections of their intersection, and m,m is a 
point on that line, mn—m'n' is a ray of light through m,m\ 
and it pierces the horizontal plane at n, outside of both planes. 
Hence, when ~Pb— P'a' is regarded as the edge of the salient die- 
dral angle, indicated by the portions, QP and RP, of the traces, 
it is an edge of shade for the direction, mn—m'n' of the light. 



SHADES AND SHADOWS. 9 

If PQP' as now shown, and that part only of PBP', extending 
in front of the edge P&— PV, are considered as the surfaces 
exposed to the light, the left side of each plane is then the one 
considered, and both of these surfaces are in the light. 

The next problem affords a more practical example of th 
general constructions just given. 

28. Second. — The lines of shade on the abacus and pillar. — See 
PI. I., Fig. 4, where the projections of a square pillar and its 
abacus, resting on the horizontal plane, are given, together with 
the direction of the light, Cm — D W. 

Here it is evident that t — t"t' ; C — CD'; the upper edges 
CE and E A ; the vertical edges at A and a ; AB — A ; B ; , and 
BC — B'C, are the edges which constitute the compound line of 
shade of the given body. They are therefore (9 — 11) the edges 
which cast shadows on the pillar, or on other surfaces ; and they 
are identical with the lines, which, in geometrical line-drawings, 
are inked as heavy lines. 



Problem IT. 

To determine the edges of shade on a pyramid, whose axis is oblique 
to both planes of projection. 

Construction. — PI. IV., Fig. 15. To make the problem quite 
definite, let G^L" be the ground line of an auxiliary vertical 
plane, parallel to the axis of the pyramid, which we will suppose 
to be a triangular one, in order to avoid needless repetition of the 
same operations of construction. Let the plane of the base be 
perpendicular to the axis, and let a"c" be its trace on the auxi- 
liary vertical plane. Let ABC be the true size and form of this 
base, seen when its plane is revolved about a"c", as its trace, into 
the auxiliary vertical plane, and let D be the projection of the 
axis on the plane of the base. Then by counter-revolution, A,B 
and C return in arcs, projected in A« /; , etc., perpendicular to 
a"c", to the points a", b and c ' ; and D, the foot of the axis, to 
d" . Then make d"v'\ perpendicular to a'V, equal to the axis 
of the pyramid, and join v 1 ' with a", b" and c" to complete the 
auxiliary projection. 

To make the horizontal projection, make aa", etc., equal to 
Az", etc., and perpendicular to the ground line Gr"L" ; and v — 



JO GENERAL PROBLEMS. 

abc will be the horizontal projection of the pyramid. The vertical 
projection, v' — a'b'c', is found, as in all similar cases, by pro- 
jecting up v — b, etc., in perpendiculars to GL, and by making 
the heights of the points v , — b\ etc., above the ground line GrL, 
equal to their heights v"h, b"h, etc. above Gr"L". 

Now, to test any edge of the pyramid, to see if it be an ed&e 
of shade, or not. 

First Method. — Let the edge vb — v'b' be taken. Find the 
traces, on one plane of projection, of the planes of the faces of 
which the given edge is the intersection. Then draw a rav 
through any point of that edge, and if it meets the plane of the 
traces outside of those traces, when the edge belongs to a salient 
angle turned towards the light, the given edge is a line of shade. 
Accordingly, vb — v'b' pierces the horizontal plane at m, and 
cb — c'b' pierces it at n, giving win, as the horizontal trace of the 
plane of the face vbc — v'b'c '. Again, va — v'a' pierces the hori- 
zontal plane at a, giving ma as the horizontal trace of the plane 
of the face vab — v'a'b'. Now a ray, bp — b'p', through any point 
bb r of the intersection, vm — v'm, of these planes, pierces the 
horizontal plane at^>, outside of both traces, ma and mn. But 
the parts of these planes which are the faces of the pyramid, esti- 
mated from the common edge, vb — vb', have ma and mO for 
their horizontal traces. Hence the direction of the light evidently 
strikes only the interior face of each of these planes, that is, the 
face towards the axis of the pyramid ; so that the exterior sur- 
faces vbc and vba are both in the dark, and vb — v'b' is not a line 
of shade. 

If now we find O, the intersection of the edge, vc — v'c' with 
the horizontal plane, aO is the horizontal trace of the plane of 
the face vca — v'c r a\ and vcO is a salient edge, similar to Fb — 
PV in Fig. 14, so that the ray, cr — cV, by piercing the hori- 
zontal plane outside the traces On and Oa, shows that vc — v'c' 
is an edge of shade, which, indeed, was known as soon as it was 
found that vb — v'b' was not such an edge. 

Second Method. — Find the shadows of all the edges of the 
pyramid on either plane of projection. Then, if, for example, 
the shadow of vc — v'c' upon the horizontal plane is further from 
the pyramid, in the direction of the light, than that of vb — v'b', 
it would show that vc — v'c', only, of those two edges, really 
sast a shadow. Hence it would be shown that vc — v'c was an 
edge of shade (26). That is, we have this principle: TIios* 



SHADES AND SHADOWS. 11 

edges, whose shadows are the boundaries of the real shadow of a body, 
are the edges of shade of that body. 

Remark. — The next problem, if not the preliminary general 
principles already given, will enable the student to construct this 
second method. 

§ II. Shadows. 

29. Here observe — First : That whenever a plane is perpen- 
dicular to either plane of projection, its entire surface, and all 
contained in it, is projected in the trace of the plane on that 
plane of projection. Second: Having a point, p, which casts a 
shadow on a plane, the projections of that shadow must be in 
the projections of the ray (4) through p. 

30. From the two foregoing principles we have the following 
simple general method (22) of finding points of shadow on the 
planes of projection, or on any planes which are perpendicular to 
them. Find, by inspection, the intersection of the linear projection 
of the plane receiving the shadow, with the corresponding projection 
of the ray through any point casting a shadow upon it. This will 
be one projection of the point of shadow sought. Its other projection 
will be the intersection of the projecting line of the point found, with 
the other projection of the same ray. 

It is only necessary here to remember, that the linear projection 
of both planes of projection, is the ground line. 



Pkoblem III. 

To find the shadow of a straight line, on the horizontal plane of 

projection. 

Principles. — The shadow of a straight line on a plane, is the 
mtersection of that plane with a plane of rays passed through 
the given line (5-6). This shadow will therefore be a straight 
line. But two points determine a straight line, and two parallel 
lines determine a plane. Hence if, through any two points of 
the given line, we pass rays, their intersection with the given 
plane will determine the shadow of the given line. Also, the 
point in which the given line meets the given plane is a point of 
the required shadow, for the line and its shadow are in the same 
plane (of rays) and therefore intersect. 



12 GENERAL PROBLEMS. 

Construction. — PI. L, Fig. 2. Let ab — a'b' be the given line, 
and r — r the direction of the light. By (30) the ra}^ ac — a'c 
drawn through any assumed point, as a, a' of the given line, 
pierces the horizontal plane in the point whose vertical pro- 
jection is c', and whose horizontal projection is therefore c. 
Likewise the ray bd — b'd' pierces the horizontal plane at d',d 
(naming that projection first which is found first). Hence cd is 
the shadow of db — a'b' on the horizontal plane. Also e,e', the 
point in which ab — a'b' pierces the horizontal plane, is a point 
of its shadow on that plane, and hence will be found in dc pro- 
duced. 

Remark. — In taking a strict view of the principles of this solu- 
tion, and in similar cases, it should be observed that the rays, as 
ah— a'b', are to be regarded as cut from the plane of rays (6) 
passed through the line, by secant planes of rays, which may 
have any position, but are most readily conceived of as perpen- 
dicular to one of the planes of projection. 

31. When a line is perpendicular to a plane, its shadoiv, on thai 
plane, is in the direction of the projection of the light on that plane y 
for the plane containing the line and a ray through any point of 
it, is a plane of rays (5), perpendicular to the given plane, and is 
therefore a projecting plane of all rays intersecting the line. 
Hence its trace on the given plane, is both the shadow of the line 
(6), and the projection of a ray, upon that plane. 

This principle is of very frequent application to the shadows 
of vertical lines upon the horizontal plane, and of co-vertical lines 
(that is, of lines perpendicular to the vertical plane) upon the 
vertical plane. 

Problem IV. 

To construct the shadow of a square abacus upon a square pillar, 
and of both, on the horizontal plane of projection. 

[There being no new principles in this problem, we pass at 
once to its graphical solution as a new illustration of the method 
of (SO) referring to (28) for the line of shade.] 

Construction. — PI. I, Fig. 4. Let the body be seen obliquely 
as shown on the vertical plane, and let Cm— T)'m' be the direc- 
tion of the light. The point d,d\ found by drawing the ray qd, 
and projecting d at d\ casts the point of shadow q,q' on the ver- 



pin. 




SHADES AND SHADOWS. 13 

tical edge at q ; q' being found at the intersection of this edge 
with the vertical projection, d'q', of the ray through d,df. From 
q' the shadow q'c r is parallel to the line de — d r e\ which casts it, 
since that line is parallel to the vertical face, qc, of the pillar (6). 
The ray through B,B' pierces the front of the pillar at g, whose 
vertical projection is in the vertical projection, B'g\ of the ray, at 
g f . Hence c'g' is the shadow of the portion, eB — e'B', of the 
edge, AB— A'B', of the abacus, upon the front of the pillar. 
g't', parallel to B'C, is the shadow of the portion B/— Wf of the 
edge BO — B'C, on the parallel front of the pillar. The ray 
through fif, passing beyond the pillar, pierces the horizontal 
plane (30) at h\ h, and hi is the shadow of the portion, t—t't", of 
the edge of the pillar, which casts a shadow on the horizontal 
plane. The portion fO—f / C / of BO — B / C / casts the shadow hk, 
beginning at h and limited at k\k by the ray Ck— Q>'k r . The 
edge C — O'D' casts the shadow km, limited by the intersection of 
the rays Qk—G r ¥ and Cm-DV with the horizontal plane. 
Similarly, CE has raw, equal and parallel to itself, for its shadow ; 
E A has, likewise, no for its shadow ; the vertical edge at A deter- 
mines the shadow op, found as km was ; pr is the shadow of 
Ad—A'd', and rq is the shadow of the vertical edge of the pillar 
at q. 

Problem Y. 

Having the projections of a ray of light, upon two vertical planes at 
right angles to each other ; to find its projections on those planes, 
after revolving it 90° about a vertical axis. 

Principles. — When we view any object in reference to a 
given vertical plane, the direction of the light is regarded as 
fixed ; and it is therefore fixed in projection, while we are find- 
ing the shadow of the object on that plane. When, however, 
we turn and face the object in a different direction, the light 
may be supposed either to turn with us, so as to come in the 
same relative direction with respect to the new projection that it 
did with respect to the old ; or it may be supposed to remain 
absolutely fixed, as it is practically in Nature for any short 
interval. A little reflection on tbe latter supposition, will show 
that, in certain aspects of an object, only its shade could be seen ; 
hence it is sometimes, if not generally, desirable to adopt the first 
view, viz. that the light turns with us, so as to come in the same 



14 GENERAL PROBLEMS. 

relative direction with respect to each new vertical plane of pro- 
jection. This first view is adopted in the following — 

Construction. — See PL II., Fig. 6. Here let the given projec- 
tions of the light, on the two vertical planes, be R'L' and R^L", 
and let qpx' indicate the real position of the vertical plane of the 
left hand projection. By an obvious construction, we find EL, 
the horizontal projection of the given ray, WL" — RT/. Then, 
as the observer turns 90° to face the plane qp, the light turns 
with him, an equal amount, about any vertical axis, as at L, and 
so appears in horizontal projection, as at rL, perpendicular to RL. 
The vertical projections of rL are, obviously, r'\J and r"\J' '. 

We see now, by inspection, that, as the vertical planes, and 
also the two positions of the light, as seen at RL and rL, are at 
right angles to each other, pt—sm, and pq—mn. That is, x'y f 
being the same for all the vertical projections of the ray, the new 
vertical projections, rT/ and r"I/', make the same angle with the 
ground line as RT/ and R"L" do. So that, in practice, it is 
only necessary to draw r'U parallel to R'T", and r"\J' equally 
inclined with RT', — but in an opposite direction — in order to 
obtain the projections of the light as it proceeds in viewing the 
side or left hand elevation. 

When the projections, rZ— rT, of the light make the conven- 
tional angle of 45° with the ground line, as in PI. II., Fig. 7, 
the vertical projection, r""l'"\ of it on the right hand plane, as 
it comes when the observer has turned, from viewing that plane, 
to view the left hand one, simply inclines at 45° to the ground 
line in the opposite sense, from rT. 



Problem VI. 

To find the projections of the shadows on the parts of a timber fram- 
ing, shown in two elevations, on two vertical planes at right angles 
to each other. 

Principles. — The chief new principle in this problem is, that a 
point of shadow may exist, geometrically, on a material surface 
produced ; though, physically, only on the real portion of that 
surface. 

Construction. — PI. II., Fig. 8. MJ — J'J" is a horizontal 
timber, resting on the transverse timber AJ— ATTC, and 
)n the lateral braces hh—g'y and hh—nu. These braces arc 



SHADES AND SHADOWS. 15 

framed into the post LH — A'H/, in front .of which is the front 
brace GP — GT 7 . The left hand elevation is supposed to be 
made on a vertical plane at right angles to the plane of the paper. 
Taking the light in this practical problem in the conventional di- 
rection (21) GL — G-T/ are the projections of a ray, as it proceeds 
in viewing the object as seen on the right hand plane of projection. 

This being established, we are prepared to construct the sha- 
dows in Fig. 8. The point A, A 7 casts a shadow on the front 
brace at a, whose other projection, a', is the intersection of the 
projecting line, a—a\ with the other projection, AV, of the ray 
Aa. A — A'B', being parallel to the face of this brace, its shadow, 
a f a", is parallel to A — A'W. Next, c,c' and L,I/, the shadows 
of the points a,a" and G,G' in the front right hand edge of the 
brace, are found precisely as a,a' was, and they determine LV, the 
position of the imaginary shadow (in this figure) of this edge on 
the plane of the front of the post produced. The shadow of the 
back left hand edge, which pierces the post at P,P 7 , begins, 
therefore, at P,P' and is parallel to LV, as seen at P'K. The 
front of the lateral brace, g'y, being parallel to the front of the 
post, the shadow of Ga — GV" upon it, will be parallel to L V. 
One point of this shadow is hf>", found by drawing the ray ab — 
a"h" ; hence h"f, parallel to LV, is the required shadow. The 
point A, B' casts the shadow hfi\ found like the preceding, on 
the plane of the front of the lateral brace. Then h'e\ parallel to 
B'C', and limited by the ray through 0,0', is the shadow of AC — 
B'C' on the same plane. The portion of shadow on the brace, 
bounded by CV, is cast by the edge, CJ — C, of the transverse 
timber, and Vb" is the shadow of A — A'B' on the plane of the 
front of this brace. Finally, for the right hand elevation, the edge 
J — J' J" casts the shadow n — y on the lateral braces, found by 
projecting over k'\ where a ray through any point of this edge 
meets, the plane of the front of these braces. 

Proceeding to the side elevation, two planes of rays, containing 
the edges whose projections are J and M, will include between 
them the shadow in space (10) of the longitudinal timber JM — 
J' J". Their traces on the side of the transverse timber, will 
therefore bound the shadow on that timber. A portion of one 
of these traces is seen at Zh" . The edge AB — B / casts a shadow 
on the side of the front brace at T ; , found by drawing the ray 
BT' (Prob. V.). According to the construction of all the other 
points of shadow in this problem, T, its other projection, is at 



16 GENERAL PROBLEMS. 

the intersection of the projecting line, T'T, with the other pro- 
iection, AT, of the ray through A, B', as it comes — according to 
previous explanation (Prob. V.) — when viewing the framing as 
seen in the left hand projection. AB being parallel to the side 
of the front brace, its shadow, tv, through T, is parallel to itself 
Lastly, the brace hk" — g'y casts a shadow on the side of the post. 
A little consideration of a model will show, that, when the brace 
makes a less angle with the horizontal plane than the light does, 
its left hand front edge and right hand back edge, as seen on the 
left hand elevation, will cast shadows. When the former angle 
is the greater of the two, the right hand front edge and left hand 
back edge will cast shadows. When the two above angles are 
equal, the front and back planes of the brace will be planes of 
rays, and their traces, kk'" and gh, produced, will bound the 
shadow of the brace. 

In the present problem, the brace makes a less angle with the 
•horizontal plane than the light does, hence the edges, k'y — kk" 
and gf—hp, cast shadows on the post. The shadow of the former 
edge begins at k, and of the latter at A, these being the points 
where these edges pierce the post. Assume any point F, F' and 
draw the ray F / H / — FH, which, according to the previous con- 
structions, pierces the plane of the side of the post atH^H. H& 
is therefore the shadow of kk" ; and sh parallel to it is the shadow 
of hp. 



Problem VII. 
To construct the shadow of a circle on the vertical plane of projection. 

Principles. — The only essential difference between this and the 
previous problems is, that, in case of a curve, a plane of rays can 
generally contain but one or two points of it, while it may con- 
tain the whole of a straight line. But we have seen that the 
shadow of a straight line is practically determined by the sha- 
dows of two of* its points. Now the shadow of a point is found 
in the same way, whether that point be on a straight line or on 
a curve ; hence shadows of straight lines, and shadows of curved 
lines, are not here distinguished as belonging to distinct classes 
of problems. 

According, therefore, to the method of (30), again, we have 
the following — 




13 



PI. Ill 



W 



B' 



^ 




PI III. 




Ill' r i u a 1 f b '• A j 







SHADES AND SHADOWS. 17 

Construction. — PI. III., Fig. 10. Let the plane of the circle 
be perpendicular to the ground line. Then the equal lines, AB 
and CT> 7 , perpendicular to the ground line, will be the projec- 
tions of the circle. Let Aa — AV be the projections of a ray of 
light. This raj meets the vertical plane at a 7 , the shadow, there- 
fore, of A, A 7 , the foremost point of the horizontal diameter of 
the circle. The ray B6 — A!b' determines the shadow of the 
point B,A 7 , at b' ; the rays Cc — O'c' and Cc — T)'d' determine the 
points of shadow, c 7 and d\ of the extremities of the vertical 
diameter — D 7 C 7 . Assuming either projection, as E, of any 
other two points on the circle, their other projections must be 
constructed by some one of the methods of (Prob. XL., Des. 
Geom.), E is the horizontal projection, as indicated by the con- 
struction, of the two points E 7 and F 7 . Each of these, again, is 
the vertical projection of two points, of which the two foremost, 
one in each pair, are horizontally projected at Gr — found by 
making CGr=CE. These four points, Gr,E 7 ; G,F ; E,E 7 ; and 
E,F 7 cast shadows at e 7 , g\ n\ and/ 7 , respectively. By joining 
the points thus found, and tinting the inclosed figure, we shall 
have the required shadow of the circle on the vertical plane. 

Remarks. — a. The lines a — a' ; b — V ; C 7 — c 7 , and D 7 — d\ are 
tangents to the shadow, which is an ellipse. Also, a' — b\ and 
c 7 — d\ are conjugate diameters, from which the axes can be 
found, if desired, as commonly shown among problems on the 
ellipse. 

b. Observing that in PI. L, Fig. 4, Jem is the shadow of 
a vertical edge at C on the horizontal plane of projec- 
tion, and that in Plate III., Fig. 10, a'V is the shadow of the 
diameter AB — A 7 on the vertical plane, we conclude from 
inspection, what was proved in (31), that whenever a line is 
perpendicular to either plane of projection, its shadow on that 
plane is in the direction of the projection of a ray of light on that 
plane. 

33. PI. III., Fig. 11, illustrates the forms of the shadows of a 
circle, whose plane is oblique to the direction of the light, upon 
several planes variously inclined to the plane of the circle. 

AB represents a vertical projection of a horizontal circle, and 
ABGC the vertical projection of an oblique cylinder of rays 
having the circle AB for its base, and parallel to the vertical 
plane of projection. The traces drawn through C, are the 
traces of a number of planes, perpendicular to the vertical plane 

2 



18 GENERAL PROBLEMS. 

of projection, and whose intersections with the cylinder of rays 
are the shadows of the circle AB. 

DO is a circular shadow, equal to AB, its plane, DC, being 
parallel to AB. CF is an elliptical shadow less than the circle 
AB, and whose transverse axis, projected at the point f, is equal 
to the diameter of AB. CF being a plane of right section, con- 
tains the least shadow of AB. CH is an elliptical shadow 
greater than AB, CH, its transverse axis being greater than AB, 
and h, its conjugate axis, being equal to AB. Between CF, the 
least, and CH, the greater shadow, there must be one equal to 
the circle AB. This is found by making GCF— FCD when 
CG=CD, and as all the diameters h, g t f, etc., are equal, CG is 
another circular shadow, called the subcontrary section of the 
cylinder, to distinguish it from the parallel section CD. Finally, 
all shadows between the circular ones CD and CGr, are ellipses, 
less than, and all exterior to these are ellipses greater than the 
circle AB. 

34. The foregoing problems being sufficient to illustrate the 
method of (30) we have the following general method for the 
more general case in which the plane receiving the shadow is 
not perpendicular to either plane of projection. Pass planes of 
rays, each of which will cut a point, p, from the line L, casting the 
shadoiv, and a line, k, from the plane, Q, receiving the shadow. 
The point in which a ray through the point, p, meets the trace, k, 
will be a point of the shadoiv of L on Q. 

Remarks. — a. When L is a straight line, the plane of rays 
may always contain it, instead of merely cutting points from it, 
and then the trace of the plane of rays on the plane Q, will be 
the shadow of L. 

b. The trace just mentioned, must, however, be constructed 
b} r the above method, as seen in the line cd, Prob. III. 'Hence, 
practically, planes of rays are not immediately passed through 
straight lines ; i. e. without construction, unless those lines are 
perpendicular to a plane of projection ; in which case the planes 
of rays containing them will be so also, and their traxs will be 
at once known. (29.) 



SHADES AND SHADOWS. 39 

Problem VIII. 

To construct the shadoiv of a straight line on a plane whose traces 
are parallel to the ground line. 

Construction. — PL I., Fig. 3. as — a f s f is the given line, and 
MK and M'K' are the traces of the given plane. OTO' is a 
vertical auxiliary plane which contains the given line, as — a f s', 
and intersects the given plane in the line nT — n'O'. Hence, 
from Des. Geom. (Prob. VI.), the given line intersects the given 
plane at s\s, which (Prob. III.) is one point of the required 
shadow. LJI/ is another vertical auxiliary plane, containing 
the ray ah — a'b\ drawn through the point a,a f of the given line 
as — a/s' '. This plane intersects the given plane in the line LJ — 
&T/, hence the ray through a,a' intersects the given plane at 
b'jbj which by (34) is therefore another point of the required 
shadow. Hence bs — 5V is the required shadow. 

Remarks. — a. The auxiliary planes might have been placed 
perpendicular to the vertical plane of projection. In that case, 
the horizontal projections b and 5 of the points of the shadow 
would have been found first. 

b. Let this problem be solved thus, and also when the given 
plane has any oblique position. 

Problem IX. 

To find the shadoiv of a chimney -, situated on the end portion of a 
hipped roof upon the end and side roofs. 

Principles. — It only need be noticed here, first, that as most 
of the edges of the chimney are perpendicular to one, or the 
other, of the planes of projection, the auxiliary planes of rays 
may all contain edges of the chimney, and be also perpendicular 
to one of the planes of projection; and second, that when a 
plane, P, is perpendicular to either plane of projection, all 
lines in that plane, P, will be projected on the plane of projec- 
tion, to which it is perpendicular, in its trace on that plane 
(29). 

Construction.— PL I., Fig. 5. TJRR'T— K / T / is the horizon- 
tal plane of the eave-trough, intersected by the roof in the lines 



20 GENERAL PROBLEMS. 

TE ; EG, and GU. GTU— G T T is the front root; ELG— Ga 
the end roof, perpendicular to the vertical plane of projection, 
and EIT — G-'I'T' is the back roof; xio is the chimney-flue, 
due a horizontal section of its outside surface, and ABD — A 'CD' 
is its abacus. 

The shadow of the abacus on the front of the chimney, is 
found as in Prob. IV. 

To find the shadows on the roof: First, the shadow on the 
front roo£ CY' is the vertical trace of a plane of rays through 
BC — C . and perpendicular to the vertical plane of projection. 
According to principle second, OT 7 is the vertical projection of 
its trace on the front roof. Y — Y"Y is its trace on the 
horizontal plane of the eave-trough ; hence, projecting O' at O 
— t have OY and OY", for the horizontal projections of its 
traces on the front and back roofs. Drawing the ray BA — 
whose vertical projection is C'Y' — and projecting h at h f , the 
point h, N is the shadow of B,C f , and hO — h'0\ the shadow of 
BC — C on the front roof. Drawing a similar plane of rays, 
B L . its ::: :rs on the roof are F L — FL", parallel to CKY'— 
OY", a line from L parallel to YO, and the short line, paral- 
lel to GE. across the end roof at F. The ray B^ — BT/ meets 
the trace FL" at g^f, giving Jig — Ji'g for the shadow of B — B'C. 
From g$ the shadow of AB — A'E', on the parallel front roofj 
is the parallel line, gf—gf. Drawing the ray /K— /'K', we 
find the portion BK — B Tv' whose shadow is gf—g'f. 

Second— -The shadow on the end roof, ec is the trace, on this 
roof, of a vertical plane of rays through the vertical edge, e — 
e'Y, of the chimney. This plane cuts the edge, AB — AB', of 
the abacus in the point a,a\ through which, drawing the ray 

c — :. and projecting c' at c. we find ec, the shadow of the 
portion e — e"e' of the edge of the chimney. Above e 7 this edge 
is in the shadow of the abacus on the chimney. Joining c,c 
and ff; cf—c'f is the shadow of dK — a'K'. Likewise, du is the 
shadow of the vertical edge at d. By projecting u at u\ and by 
drawing the vertical projection of a ray through v. . we :ould 
find the portion of the edge from d,d\ upwards, which casts the 
shadow du — du'. 

Third. — To find the shadow on the back roof and eaves plane. 
AV is the vertical trace of a plane of rays parallel to the plane 
C Y Y ". Its trace on the back roof is Wo' — :X. It intersects 
the edge, d — d'v\ of the chimney at d,v\ through which, draw- 



SHADES AND SHADOWS. 21 

m& a ray, v'o' — dp, we find p,p' the limit of the shadow of d — 
d'v' on the back roof. This point p,p' is also by construction 
the shadow of r,A' where the ray p'v' — pd meets the edge 
AD — A' of the abacus. Drawing the ray or", rr" is determined 
as the portion of this edge, which casts the shadow po — p'o ! . 
The remaining portion, Dr", casts the parallel shadow on — o', a 
part of the trace of the plane of rays AV on the plane of the 
eave-trough, and limited by the ray Dn — AV. Drawing the 
ray Yb, we find the limit of that portion of the edge BC — C 
which casts a shadow on the back roof. bG casts the equal and 
parallel shadow, Yy, on the parallel plane of the eave-trough. 
Likewise ym is equal and parallel to CD — CD 7 , and is limited 
by the rays Gy and Dm. Then mn is the shadow of D — AT)', 
which completes all the required shadows. 

Remarks. — a. The boundary of the shadow on the back roof, 
as seen in vertical projection, is fringed, to give it greater dis- 
tinctness. 

b. To locate either projection of a point on the side roofs, one 
projection of such a point being given. Let H be the horizontal 
projection of a vertical rod on the front roof. Draw Hq parallel 
to the trace, GU, of the front roof; q is vertically projected at 
q r , and g'H', parallel to G'T', is the vertical projection of qK ; 
hence H'H" is the vertical projection of the rod at H. This 
construction would be used in locating the intersection, with 
either front or back roof, of a chimney situated on either of those 
roofs. (See Des. Geom. Prob. X.) 

c. To find the shadow of the rod, H — B7H", pass a vertical 
plane of rays through it , HJ, parallel to Bg, is the horizontal 
projection of the trace of this plane on the front roof. Project 
J at J' and H'J' will be the vertical projection of the same 
trace. The ray Wh'" — HA" meets this trace at h"\h" , giving 
H.h /; — EW for the projections of the shadow of the rod. This 
construction could have been employed in finding all points of 
the shadow of the chimney, instead of making the planes of 
rays perpendicular to the vertical plane. Observe that in the 
construction given for the chimney, we find the horizontal pro- 
jections of points, as h, first; while in the construction oih"h'" 
we find h'" first. 

35. In dismissing the subject of shadows on planes, it may 
be noticed, that it is sometimes advantageous to consider the 
ray of light, itself, as making an angle of 45° with one of the 



22 GENERAL PROBLEMS. 

planes of projection. For, suppose such a ray to t>e passed 
through the uppermost point of a vertical line, and to make an 
angle of 45° with the horizontal plane. Evidently, the distance 
from the foot of the line to the intersection of the ray with the 
horizontal plane — which would be the shadow of the line — 
would then be equal to the height of the line. Hence, without 
having any vertical projection of the line, its height may be 
known from its shadow. Or, which amounts to the same thing, 
the height of a point above the horizontal plane will be known 
by the distance of its shadow from its horizontal projection. 

The same essential principle is true in more general terms. 
For, by knowing the direction of the light, we can find the rela- 
tive lengths of vertical lines and their shadows, and this relation 
being known, we can find the height of all points of an object, 
having given only its plan and shadow. The same principles 
are applicable to constructions made on the vertical plane of 
projection. To illustrate, let us take the following example : 



Problem X. 
To find the shadow of a shelf and brackets upon a vertical plane. 

Construction. — PL II., Fig. 9. Let AB, taken as the ground 
line, be the horizontal trace of the given vertical surface ; let 
CD — CT)' be the lower front edge of the shelf, and let cb — c'b' 
be one of the brackets. 

Let E'F be the vertical projection of a ray, drawn in any 
direction, relative to the ground line. It is sufficient to under- 
stand that this ray makes an angle of 45° with the vertical 
plane. Then the shadow, c'd', of the edge en — c' will be parallel 
to E'F and equal to en. Next, d'e' can be drawn, parallel tc 
ca — cV, and limited by the ray aV, and the shadow of the 
unshaded part of a — a'b' will be the parallel line e'k'. 

Likewise, taking any point ff, in the edge CD — C'D' of the 
shelf, make the ray fh' equal to fg, then h'k\ parallel to CD', 
will be the shadow of CD— C'D 7 . 

Remark. — Though not necessary, the horizontal projection of 
the ray may be found as follows : Make GrA - E'F, then will 
GF — E'F be the two projections of a ray which makes an angle 
of 45° with the vertical plane of projection. 



SHADES AND SHADOWS. 23 



CLASS II. 

SHADES AND SHADOWS ON SINGLE CURVED SURFACES, L2f 

GENERAL. 

SECTION I. 
On Developable Single Curved Surfaces. 

§ 1. — Shades. 

36. The line of shade on a developable surface, consists of 
those rectilinear elements along which planes of rays are tangent 
to the surface, since such a line would evidently separate the 
illuminated from the dark portion of the surface. Since the 
tangent planes are planes of rays, they may be regarded as 
parallel to a given ray, in space. Hence the general method for 
the construction of the elements of shade on a developable sur- 
face, resolves itself into the operations of (Des. Greom. Probs. 
53, &c.) " To construct a plane, parallel to a given line, and tan- 
gent to a developable single curved surface" Since two such planes 
can generally be found in any case, there will usually be two 
opposite elements of shade. 

37. When the indefinite surface is limited by one or more 
plane intersections, taken as bases, the complete line of shade 
will consist of the elements of shade on the convex surface, toge- 
ther with those portions of the circumferences of the bases which 
divide light from dark portions of the entire body. 

38. If a plane of rays be tangent to a developable surface 
along one of its elements, then any secant plane will cut from the 
surface a curve, and from the plane of rays, a line, tangent to the 
curve at one point of the element of shade. Moreover, if the secant 
plane be perpendicular to the tangent plane of rays, the line cut 
by it from the latter will evidently be the projection of the light 
upon the secant plane, since the tangent plane, when thus per- 
pendicular to the secant plane, becomes a projecting plane of 
rays upon the latter plane. 



24 GENERAL PROBLEMS. 

Problem XL 
To construct the elements of shade on any cylinder or cone. 

Constructions.- — First. — To construct the element of shade on a 
horizontal right semi- cylinder. PI. IY., Fig. 16. Let Fb be the 
trace of one of two vertical planes of projection — upon the plane 
of the paper — taken as the second vertical plane, and at right 
angles to the former one. Let the former plane be perpendicular 
to the common axis of the cylinder, and cylindrical abacus, shown 
in the figure, and let the latter plane contain the axis of the 
cylinder. The projections will then be as in the figure. Let 
De — DV be the projections of a luminous ray. De is also the 
trace, on the right hand plane, of a plane of rays perpendicular 
to that plane ; e is therefore the right hand elevation of its 
element of contact with the cylinder, that is, of the element of 
shade of the cylinder. We" is therefore the left, or linear elevation 
of the same element. 

Likewise, the parallel tangent plane of rays at Gr, determines 
the element of shade, Gr — Gr^Gr', on the abacus. 

/Second. — To construct the elements of shade on a cylinder whose 
axis is oblique to both planes of projection. PI. III., Fig. 13. ABF 
— A'G' is the lower base of the cylinder. Ibe — I'J' is its upper 
base, Qb and M.g are its extreme elements, as seen in horizontal 
projection, and A'l' and Gr'J', the extreme elements as seen in 
vertical projection. The axis of the cylinder pierces the hori- 
zontal plane at K, the centre of the lower base. The ray dL — 
a'L', through a point a,a' of the axis, pierces the same plane at 
L, hence KL is the horizontal trace of a plane containing the axis 
and a ray. From (5) such a plane is a plane of rays through 
the axis. The tangent planes of rays will be parallel to the one 
just found, hence their horizontal traces will be parallel to KL. 
BB r/ and FF" are these traces, and their points of tangency, B 
and F, with the lower base, are where their elements of contact 
— the elements of shade — pierce the horizontal plane. Hence Bw 
— BV, and Ff—Ff, are the required elements of shade; one 
projection of each being visible. 

Third. — To construct the elements of shade on a cylinder whose 
axis is horizontal, but oblique to the vertical plane of projection. PL 
IY., Fig. 17. In order to construct the vertical projection of a 
cylinder, thus seen obliquely, it is necessary to have, first, its 



PI. IV 




N. IV 




SHADES AND SHADOWS. 25 

projection on a plane perpendicular to its axis. Then let NZH 
be the horizontal projection of the cylinder, and take CC", 
perpendicular to its axis, for the ground line of a new vertical 
plane, parallel to the base NZ. This base will then have the 
circle WO"b" for its auxiliary vertical projection, and that of the 
cylinder. Now, any point as Z ; , in the principal elevation, is in 
a projecting line, ZZ', and at a height, Z'O', from the ground 
line, equal to the height of the same point, as seen at Z", above 
the ground line 0"G". 

Having thus found the projections of the cylinder, we may 
assume any two projections of a luminous ray, and find the third. 
Let FL — FT/ be the assumed projections of a ray. L,I/ is pro- 
jected on the auxiliary plane at I/ 7 ; — I/, and L" being at equal 
heights above their respective ground lines. Likewise Gr" and 
F', both projected from F, are at equal heights above their 
respective ground lines. GK1/' is then the auxiliary projection 
of the ray. Hence also T"H" and M"J", parallel to (x"L", are 
the traces, on the auxiliary vertical plane, of tangent planes of 
rays perpendicular to that plane. Hence H" — HI and Z" — KJ 
are two projections of the two elements of shade of the cylinder. 
KJ is again vertically projected at K'J' at a height, equal to 
J'V, above the ground line CO', or by projecting J into the 
principal vertical projection of the base at J', and drawing J'K' 
parallel to the ground line. The former construction is obviously 
more accurate in practice. 

Otherwise : omitting all the operations of the last paragraph, 
the ray, NO — WO' ', may be projected into the plane, NZ, of the 
base of the cylinder, by projecting 0,0' at o'\o'" . Then, con- 
sidering that N,N' is its own projection on the plane NZ, we 
have !&o" — -Wo'" to represent the projections of the ray on this 
plane. Hence draw a line tangent to the base, and parallel to 
Wo'" , and by the principle of (38) J' will be its point of tan* 
gency with the base WZ'TJ, and therefore a point of the element 
of shade J'K'. 

Fourth. — To construct the elements of shade on a cone. PL TIL, 
Fig. 12, and PL V., Fig. 18. 

Principles. — A plane is tangent to a cone along an element, 
and all the elements contain the vertex of the cone, hence a plane 
of rays tangent to a cone will contain the ray through its ver- 
tex. Therefore the trace of this plane on any given plane, will 
contain the intersection of that ray with this given plane, and, 



26 GENERAL PROBLEMS. 

by (38) the point of contact of this trace with the section of tnt 
cone, made by the given plane, will be a point of the element of 
shade. 

Constructions. — In PL III., Fig. 12, draw a ray, YE — Y'E', 
from the vertex, V, V, of the cone. It pierces the horizontal 
plane, which is here the plane of the cone's base, at E. Then 
ET is the horizontal trace of one of the two tangent planes which 
can be drawn, and TY — T'Y 7 is the element of shade determined 
by it. In PI. Y, Pig. 18, as the ray, YJST— Y'E 7 pierces the 
horizontal plane within the cone's base, on that plane, no lines 
can be drawn from N, tangent to the base. This indicates that 
no planes of rays can be drawn tangent to the cone. The lower 
nappe, therefore, would be wholly illuminated, but for the pre- 
sence of the upper nappe, which casts a shadow upon it. 

Fifth. — To construct the elements of shade on two intersecting 
cylinders, whose axes are at right angles to each other, in a plane 
parallel to the vertical plane of projection, one of these axes being ver- 
tical. PI. Y., Fig. 19. The elements of shade on the vertical 
cylinder are found at once, by drawing two vertical planes of 
rays, tangent to that cylinder. Af is the horizontal trace of 
such a plane, coinciding with the horizontal projection of a ray, 
and giving the element of shade A — A! A!'. 

The elements of shade on the horizontal cylinder, are found 
by revolving the plane of either of its bases either into, or paral- 
lel to, one of the planes of projection. Let the plane uDv f be 
revolved about its vertical trace Dv\ into the vertical plane of 
projection. The centre n,n' of the base contained in this plane, 
will then revolve in the arc nn'" — n'n" , to n" . The circle, with 
n" as a centre, and radius n // T // =n / W, will then be the revolved 
position of the base uy — e'v' ', and the projection of the entire 
horizontal cylinder upon the plane uDv'. ISTow, T"k" is the 
trace on the plane of the revolved base,, of a plane of rays per- 
pendicular to that plane and tangent to the horizontal cylinder, 
T"k" being drawn in any assumed direction, if the principal pro- 
jections of a ray are not given. T", and the point Y" diametri- 
cally opposite, are then the auxiliary projections of the elements 
of shade of the horizontal cylinder. By a counter revolution, in 
which these points, considered as the extremities of these ele- 
ments of shade, revolve in the arcs T"e' — , tl'e and Y"v' — v"p, 
we find ePI — e r W for the lower front element of shade, and 
pY — v'Y' for the upper back element of shade. 



SHADES AND SHADOWS. 21 

Bemark. — If the principal projections of the light were given, 
T"k' f , instead of being assumed, would be constructed as the 
auxiliary projection of the light upon the plane wD*/, by an 
obvious construction. 



§ II. — Shadows. 

39. As in the case of planes, so in that of single curved sur- 
faces, when all their elements are parallel, the projection of the 
surface will be a line, when it is perpendicular to a plane of pro- 
jection. The projection of the perpendicular plane will, how- 
ever, be a straight line, while that of the single curved surface 
will be a curve ; viz. its curve of right section. As cylinders 
alone, among single curved surfaces, have parallel elements, they 
only can have all their elements perpendicular to the same plane, 
and hence they only can be projected in a line. 

40. For all cases of shadows upon cylinders whose axes are 
perpendicular to a plane of projection, we have, from the pre- 
ceding article, the following special method. Determine, by inspec- 
tion, the intersection, M, of a ray through any point, P, of the line 
casting the shadow, with the linear projection of the cylinder. This 
will be one projection of the shadow of P. Its other projection vjUI 
be the intersection of the line 'perpendicular to the ground line, through 
M, with the other projection of the ray through P. 

The two following problems illustrate this method. 



Problem XII. 

To construct the shadow of the projecting head of a cylinder upon 

the cylinder. 

PI. IV., Fig. 16. The projections of this body have been fully 
described in the first case of the preceding problem. The right- 
hand edge, G^A', of the head, above the element of shade, Gr — 
GKGr", casts the shadow required. Drawing the projection, Od, 
of a ray, on the plane containing the linear projection, bdEt, of 
the cylinder, d is at once found, as one projection of the shadow 
of 0,0' upon the cylinder. Its other projection, d\ is the inter- 
section of the perpendicular, d — d\ to the trace (ground line) Yb, 
with the other projection, G'd\ of the ray through 0,(7. The 



28 GENERAL PROBLEMS. 

other points of shadow are found in exactly the same way, as ia 
evident by inspection. The tangent ray, De — DV, determines 
that point of shadow, e,e\ in which the shadow is lost in the 
element of shade, e — e'e". 



Problem XIII. 

To construct the shadows of the edges of shade of a cross upon a 
cylinder ', both bodies being seen obliquely. 

Principles. — PI. IY., Fig. 17. The three projections of the 
cross and cylinder, of the ray of light, and of the elements of 
shade of the cylinder, having been already constructed, as 
explained in Prob. XI. (Third), we may proceed at once with 
the construction of the shadow, after having, first, determined, 
by inspection, the edges of shade of the cross, and, second, the 
limits of that part of the body of the cross, whose edges of shade 
cast shadows on the cylinder. 

The edges of shade must, evidently, here be determined with 
reference to a single direction of the luminous ray in space ; that 
is, we must find the projections of those edges, on the three 
planes of projection, which in space really cast shadows, when 
the light comes in the single direction projected in FL — FT/ — 
GK'L". These edges are, in the present case, B'T)"— BD— B'D' ; 
the one diagonally opposite, through YY" ; then c" — c'"Y — c'P', 
and the one diagonally opposite, G" — FGr — F'Gr' ; also X^Gr" — 
XGr — X'Gr' ; its diagonally opposite edge, c"w" — c'"w ; next 
X"c" — XP — XT 7 ; and its diagonally opposite edge, ¥w — G'V, 
and the horizontal edge at ~D",D,D'. 

The limits of that portion of the cross, which casts a shadow 
on the cylinder, are found by drawing the two tangent planes of 
rays, whose traces are M^J" and T^H", and which determine 
the elements of shade on the cylinder (Prob. XI. Third). These 
planes contain those points of the cross, which cast shadows on 
the elements of shade, where the shadow disappears from the. 
cylinder. 

Construction. — The }_oint M'^MjM', cut from the cross by the 
plane M^J", casts a shadow on the element of shade, at a point 
whose auxiliary projection is J", whose horizontal projection 
may be found by projecting Z" into the horizontal projection, 
Ma, of the ray through M^M, at a ; and whose vertical pro- 



SHADES AND SHADOWS. 29 

jection may then be found by projecting a into the veriical pro- 
jection, MV, of the same ray, at a'. Otherwise : we may con- 
struct the vertical projection, J'K', of the element of shade, on 
which the shadow of M // ,M / is known to fall, as explained in 
(Prob. XL Third). Then the intersection, a', of this element 
with the ray, vertically projected in MV, is the vertical projec- 
tion of the required point of shadow, which may be then pro- 
jected into the horizontal projection, Ma, of the same ray, 
at a. 

Thus every point of shadow may be found, first in horizontal 
projection, and then projected up ; or first in vertical projection, 
and then projected down. Also, a may be projected up into 
J'K', instead of into MV ; or, a' being first found, as above, it 
may be projected down into JK, to find a. Indeed this is. prac- 
tically, the most exact construction. 

As all the other points of the shadow are found precisely as 
just described, we shall only point out those, which, in the 
present figure, it is desirable to find. This will assist in solving 
the problem, when the relative position of the bodies, the light, 
and the planes of projection, are slightly changed. 

The shadow on the foremost element, b" — v~N — v'N" 7 , is cast 
by the point, K", K,B/, in the ray b"~EL", and falls at b",b,b r . 

In the figure, the cross is made tangent to the cylinder along 
the edge c" — c /7/ P, which is therefore its own shadow. The 
highest point, S", of the edge B'OC", which casts a shadow, is found 
by drawing the ray c f, 'S ;/ . Its shadow is c h \c,c' ' . The frag- 
ment, S"X", casts the shadow, cT, on the front of the short arm 
of the cross. 

The shadow on the highest element, is found by drawing the 
ray ^"Q", which determines the point Q'^C^Q', whose shadow 
is p'\e,e'. The edge X"c" — XP casts the shadow Pa 7 — P V ; 
the edge XGr, the shadow from d to the point just beyond e ; the 
edge GrF, the shadow from the latter point to o ; the edge through 
F and w, the shadow op, and the edge through c" and w'\ the 
shadow pc ,ff , partly invisible. The portion G"T" — UT, of LTD, 
casts the shadow gf, and the portion, V'V, of the lower left hand 
edge, the shadow hh. 

Remarks. — a. The portion of shadow cast on the horizontal 
plane in front of the cylinder is shown. It is found as in Prob. 
III. The shadow of the cylinder, and of the part of the cross 
which is beyond the plane H/'T", can be found on the horizontal 



30 GENERAL PEOBLEMS. 

plane in the same manner, and will add to the beauty of the 
figure, especially when shaded in graduated tints. 

b. If the light were to make a greater angle with the horizon- 
tal plane than the edges parallel to X'V, as seen in the auxiliary 
projection, the edges, as X" — XT, would be edges of shade. 

c. Those edges are inked as lines of shade, in the auxiliary 
elevation, which are so, in reference to the single direction of the 
luminous ray, shown in this figure. 

41. The general method for finding shadows on oblique cylin- 
ders, and other single curved surfaces, is this : Pass planes of 
rays so that each shall cut the line casting the shadow in a point P, 
and the single curved surface in a straight element, E — or in some 
other simple section. The point, S, where the ray through P meets 
the element, E, is the shadow of P upon E, and is therefore one point 
of the shadow of the given line upon the given surface. 

Remarks. — a. For the cylinder ; a plane of rays must be 
parallel to the axis, in order to intersect the cylinder in straight 
elements. 

b. For the cone ; a plane must contain the ray through the 
vertex, in order to intersect it in straight elements. 

c. The two following problems illustrate this general method. 



Problem XIY. 

To construct the shadow cast by the upper base of a hollow oblique 
cylinder upon its interior. 

Principles. — The method of (41) applied to this problem givea 
the following principles of solution. Any secant plane, parallel 
to the plane containing the axis and a ray, will cut two recti- 
linear elements from the cylinder, one of which will be towards 
the source of light. The upper extremity of this element casts a 
point of shadow on the opposite element in the same plane. The 
elements contained in the same tangent planes of rays, coincide 
in one, which is the element of shade. Hence the upper extre- 
mity of this element is also a point of shadow on the interior of 
the cylinder. 

Constructions. — PI. HI, Fig. 13. According to the foregoing 
principles,//', and u,u' ; the upper extremities of the elements 



SHADES AND SHADOWS. 31 

of shade (Prob. XI. Second) are the points where the required 
shadow on the interior begins. 

To find any other point of this shadow. The plane of rays 
whose horizontal trace, MH, is parallel to the horizontal trace, 
JSTEL, of the plane of rays through the axis, contains the elements 
Mg and lie. The point, g,g\ of the former element, casts its 
shadow on the latter element, at hji. This point may be found 
by drawing the horizontal projection, gh, of a ray, noting the 
point h, and then projecting it, either into the vertical projection, 
g'h\ of the ray, or into the vertical projection, DV, of the ele- 
ment containing it. Otherwise : the same point may be found 
by first drawing the vertical projection, g'h r , of the ray through 
g,g\ noting its intersection, h', with the vertical projection, DV, 
of the element containing it, and then projecting it either into 
gh, the horizontal projection of the same ray, or into He, the hori- 
zontal projection of the element containing h,h'. 

To find the shadow cast by any particular point, assumed in 
advance, on the upper base, as n,n' '. Draw the element, n — 1ST, 
containing this point, and through N", its intersection with the 
horizontal plane, draw the trace, NE, of a plane of rays, and 
then proceed as before, to find o y o\ the shadow of n,n\ 

Conversely, to find the shadow cast upon any element, assumed 
in advance, as Ob — Q'b f . Through the intersection, C, of this 
element, with the horizontal plane, draw the trace CR, parallel 
to 1STE, of the plane of rays which cuts from the cylinder the ele- 
ment Rr, towards the source of light ; whose upper extremity, r. 
casts the shadow, 5,5', on the assumed element bC — b'Q f . 

The lowest point, 0,0', of the shade, is evidently on the ele- 
ment, ~Ed — E'd', contained in the plane of rays through the axis, 
because the two elements contained in this plane are the farthest 
apart, being diametrically opposite. 

jRemarlcs. — a. It does not strictly belong to this problem to 
construct the shadow of the cylinder on the horizontal plane. It 
is however shown in the figure, and is bounded by BB" the 
shadow of the element of shade, Bu — B V ; the semi-circle, B'T", 
whose centre is L, and which is the shadow of udf, and FF", the 
shadow of the element of shade, ¥f—Ff. 

b. The vertical projection of the shadow on the interior of the 
cylinder is invisible. 

c. It happens in the present figure, that the horizontal pro- 
jection, fins, of the same shadow is a straight line. This shows 



32 GENERAL PROBLEMS. 

that this shadow is a plane curve, whose plane happens, in this 
instance, to be vertical. The shadow being a plane curve, and 
also the intersection of two cylinders having a common base, 
fJI — JT, — viz. the given cylinder, and the cylinder of rays 
through its upper base — we infer, what is actually true, that the 
intersections of such cjdinders, having, as they do, two common 
tangent planes, are plane curves. 

d. The last remark serves as a simple illustration of the use of 
the exact constructions of Descriptive Geometry as a means of 
research in discovering theorems, particularly of form and posi- 
tion, relating to geometrical magnitudes. 

42. In the following problem, besides an illustration of the 
general method of (41) there is an illustration of the special method 
called the method of one auxiliary shadow, which is as follows : 

Where the shadow of a line on any surface meets the intersec- 
tion of that surface with a second surface, is a point of the sha- 
dow of the given line upon that second surface. 

Thus : where the shadow of a staff, upon the ground, meets 
the intersection of the ground with a house, is where the shadow 
of the staff upon the house begins. 



Problem XY. 

To construct the shadow of the upper base of a vertical right cone, 
upon the lovjer nappe of the same cone. 

Principles. — According to (41) the upper base is the line cast- 
ing the shadow, the lower nappe is the single curved surface 
receiving the shadow, and any secant plane containing the ray 
through the vertex, and cutting the base, is a secant plane of 
rays containing two elements of the cone. The intersection of 
each of these elements with the upper base, is, in general, a point 
casting a shadow on the other element. 

Construction,— ?\. Y., Fig. 18. AGL and A'H'— V— A"H" 
are the projections of the cone, and YN — Y'E', of the ray 
through the vertex, which pierces the horizontal plane at N,E', 
and the plane of the upper base at D'",D". As all the planes of 
rays are to contain this ray, their traces on the plane of the lower 
base will all pass through N, and those on the upper base, 
through D"', as seen in horizontal projection. 



PI 




PI V 




SHADES AND SHADOWS. 33 

To find the points of shadow, which fall on the circumference of 
the lower base of the cone. These points are found by the special 
method of (42). The ray VJ — VJ 7 , through the centre of the 
upper base, pierces the horizontal plane (the plane of the lower 
base) at J'. Then a circle with J as a centre, and radius JL 
equal to V'H", that of the upper base, will be the shadow of the 
upper base on the plane of the lower base. LGr is an arc of this 
shadow, and L,I/ and G,G-', its intersections with the lower base, 
are the required points of shadow on the circumference of that 
base. Other points might be similarly found. 

Remark. — It will be observed that, by this method, we neces- 
sarily find, neither the elements nor the rays, containing these 
points of shadow ; nor, consequently, the points casting them. 
'To find the elements, connect L,I/ and G,G' with the vertex 
V,V. To find the rays, draw the projections of rays through 
these points and note their intersections with the corresponding 
projections of the upper base. The latter intersections will be 
those points of the upper base, whose shadows are L,I/ and G,G'. 

To find other points of the required shadow. These are found by 
the general method of (41). Let AM be the horizontal trace of 
any secant plane of rays. It cuts from the cone the two ele- 
ments IYB, and AH— A'V'H". The point B,B" of the former, 
in the upper base, casts a shadow on the latter at n f ,n, found by 
drawing the vertical projection Wn' of a ray, and projecting its 
intersection, n' , with A'V, at w, on AY, which last point may 
also be found as below. 

Again : let AF be the horizontal projection of the trace of a 
secant plane of rays, upon the plane of the upper base. This 
plane cuts the upper base at A, A", and the cone in the element 
FVM — M'V receiving the shadow of A, A", at a,a ; . This point 
was found by drawing the horizontal projection, Aa, of a ray, 
and projecting a into M'V at a' . The point, a, might, instead, 
have been projected into the vertical projection, A'V, of the 
same ray. It might also have been found as n,n' was. (See a,a' 
in PL IV., Fig. 17.) 

When the secant plane of rays, as at EK, is perpendicular to 
the vertical plane of projection, the vertical projections of the 
elements, and rays, contained in it, are confounded together in 
its vertical trace E'D". Hence we have not, in this case, the 
choice of methods of procedure, given above, but must find, 
first, the horizontal projection of the point of shadow contained 

3 



84 GENERAL PROBLEMS. 

in this plane. Thus : the plane ENK cuts from the cone the 
elements EY and KYD ; the latter of which determines the 
point D,!)" on the upper base, which casts a shadow on the 
former at b,b' ; found only by drawing the ray ~Db, and projecting 
b into the vertical projection, E'V, of the element EY, at b' . 

The highest point of the shadow, is in the vertical meridian 
plane of rays, CYJ. The point 0,0", cut from the upper base 
by this plane, casts a shadow on the element CY — C'V', in the 
same plane, at e,e r . 

Through the points now found, the projections of the curve of 
shadow can be sketched. In the horizontal projection, the upper 
nappe conceals the shadow. In the vertical projection, the whole 
of the front of the upper nappe, and all of the lower nappe, above 
the line of shadow n'b'di, are shaded, being visible portions of 
the darkened surface of the cone. 

Remark. — If the light had made a less angle with the hori- 
zontal plane, than that made by the elements of the cone, there 
would have been elements of shade on the cone and a shadow 
cast by the upper circle on the interior surface. It is recom- 
mended that the problem should be solved under these con- 
ditions. 

43. In the following problem occurs an illustration of the 
special method of (40), of the special method of (42), and of a new 
special method, which may be called the method by one indefinite 
rectilinear projection of the shadow, known by inspection. Thus the 
shadow of a vertical straight line, upon any surface, will be 
straight, as seen in horizontal projection. The method itself, as 
applied in finding any other projection of the shadow, is this. 
Having the intersection of the given projection of the shadow with any 
element of the surface on ivhich it falls, project that point upon the 
other projections, one or more of the same element. 

Problem XYI. 

To construct the shadow cast by the vertical cylinder (Prob. .XZ, 
Fifth) upon the horizontal cylinder, and by the horizontal cylinder 
upon the vertical cylinder. 

Principles. — The lines of each cylinder, which Cctst the sha- 
dows, are their elements of shade, and the portions of the cir- 
cumferences of their bases, which are curves of shade, as found 



SHADES AND SHADOWS. 35 

in (Prob. XL, Fifth). The shadow of the horizontal cylinder 
upon the vertical one, will be found by the special method of 
(40). The shadow of the upper base of the vertical cylinder on 
the horizontal cylinder will be found by the special method of 
(42), and the shadow of the element of shade of the vertical 
cylinder, upon the horizontal cylinder, will be found by the spe- 
cial method of (43). 

Constructions. — 1°. PI. Y. Fig. 19. To find the shadow of the 
lower front element of shade, eH — e'H, of the horizontal cylinder, 
upon the vertical cylinder. This shadow begins at tf, where this 
element of shade pierces the latter cylinder. Through any point, 
X,X', of the element of shade, pass a ray Xcc— X'a/. This ray 
meets the vertical cylinder in a point, whose horizontal projec- 
tion is x, and whose vertical projection is x\ the intersection of 
the perpendicular, x — x\ to the ground line, with the vertical pro- 
jection, XV, of the ray. In the same manner, s,s', the shadow 
of the extremity of the element of shade, is found ; also, r,r', the 
last visible point of shadow on the vertical cylinder, which is 
cast by the point, q,q\ of the left hand base, found by drawing 
the ray, rq — r'q\ backwards from r, the horizontal projection of 
the extreme left hand element of the vertical cylinder. Through 
the points now found, t'x's\ the line of shadow, cast by et—e't\ 
may be sketched ; also, *V, the shadow of the portion, eq—e'q\ 
of the base, just mentioned. 

2°. To find the shadow of the upper base of the vertical cylinder, 
upon the horizontal cylinder. Assume any horizontal plane, as 
the one whose vertical trace is W, and containing elements of 
the horizontal cylinder, in which it is supposed that points of the 
required shadow will fall. This plane cuts two elements from 
the latter cylinder, the hindmost of which is projected at YV— 
V". By making a counter-revolution, Y // v , —v // p 1 of Y", to 
v\p, we find pY, the horizontal projection of this element. OT)' 
— OD is the ray through the centre of the upper base, and it 
pierces the plane YV at D',D. A circle described with D as a 
centre, and radius equal to OT, is the shadow of the upper base 
on the auxiliary plane YV ; then c?, the intersection of this 
shadow with the element, pY, cut from the cylinder by that 
plane, is the horizontal projection of a point of the shadow of the 
upper base of the vertical cylinder upon the horizontal cylinder. 
By projecting d, at d\ into the vertical projection, YV, of the 
element in which it lies, we have both projections of this point 



36 GENERAL PROBLEMS. 

of shadow. The point b,V is found in the same manner on the 
element rrib— Wb' contained in the auxiliary horizontal plane 
whose vertical trace is B' — V . In like manner the point c,c\ on 
the highest element, n~F— E'F', is found ; the shadow of the cen- 
tre of the base OT, on the horizontal plane containing the high- 
est element, being at C^C. 

3°. To find the shadow of the front element of shade, A — A 'A", 
of the vertical cylinder, upon the horizontal cylinder. By (43) we 
have, at once, the straight line AK, coinciding with the horizontal 
trace of a vertical plane of rays through A — A' A", for the hori- 
zontal projection of this shadow. It begins at A, A', where the 
element of shade, A — A 'A", pierces the upper half of the hori- 
zontal cylinder. Any other point, as g\ of the vertical projec- 
tion, is thus found. Assume any element, as g'$\ whose projec- 
tion on the end elevation is g" . By a counter-revolution, the 
element through g" is found in horizontal projection at Jg. 
Project g, its intersection with the shadow, AK, into the vertical 
projection, J'g', of the assumed element, and g' will be determined 
as required. Likewise a,a\ on the highest element, and fif, on 
the back element of shade, produced, are found. The curve 
h'af, which, being an oblique plane section of a cylinder, is an 
ellipse, can next be sketched through the points now found. To 
find how far this shadow is real, draw the vertical projection, 
A'V, of the ray through the highest point A, A", of the element 
of shade A — A' A", note its intersection, o\ with the indefinite 
shadow h'a'f, and project o' at o. Then Ao — h'g'o' is the definite 
shadow of A — A' A" upon the horizontal cylinder. At o,o' 
begins the shadow, ocbd — o'c'b'd' , of the upper base, which is lost 
in the shade of the cylinder at d } d\ 

Remarks. — a. When, as in this case, the front element of shade 
:>f the vertical cylinder lies to the left of the point, I/, a minute 
shadow will be cast by it on the lower half of the horizontal 
cylinder near I/. When, however, the same element of shade 
lies to the right of I/, a small portion of T" — L'H' will cast a 
small shadow on the vertical cylinder. 

b. By constructing the base dx'V in end elevation, which would 
be a straight line equal to GT', and in G-T produced, and by 
constructing, in the same elevation, the element of shade 
A — A! A", both of the shadows on the horizontal cylinder could 
have been found by the simple special method of (40), applied 
very nearly as in Prob. XIII. 



PL VI 




PI. VI. 




SHADES AND SHADOWS. 37 



Pkobleh XYIL 

To find the axts of an elliptical shadow, two of whose conjugate 

diameters are known. 

Principles. — PL "VI. , Fig. 66. It has already been shown 
(Elemen. Proj. Drawing, Div. IY., Chap. Y.) and (Descrip. Geom.) 
that, if the projecting lines of an object are oblique to a plane of 
projection, surfaces parallel to that plane will be shown, still, in 
both their true form and size ; and that lines perpendicular to 
that plane will be shown as parallel lines, whose direction will 
depend on that of the oblique projecting lines. 

In case the projecting lines make an angle of 45° with the 
plane of projection, as in " Cavalier Perspective " or Cabinet 
Projection, the projection of a perpendicular to the plane of pro- 
jection will be equal to the line itself; since the line, its projec- 
tion, and fts projecting line, taken together, will form an isos- 
celes right-angled triangle. But if the projecting lines make any 
other angle than 45° with the plane of projection, the projection 
of the line will be longer or shorter than the line itself. The 
latter case is shown in Fig. 66, which is a general oblique pro- 
jection of a cube; FG, etc., being less than EF, the edge parallel 
to the plane of projection. 

To view this figure, now, with reference to the present problem, 
the ellipse mQnp is the oblique projection of the circle inscribed 
in that face of the cube which is perpendicular to its front face, 
EFLK, in the line LB.. Hence the figure also truly represents, 
in a pictorial manner, the projection of the circle MPNQ, con- 
sidered as vertical, upon the horizontal plane LRHK ; and, more- 
over, if Oo, which thus represents an oblique projecting line, is 
regarded as a ray of light, then mQnp will represent the shadow 
of MPNQ on the plane LEHK. 

Construction. — Let mn and pQ, Fig. 67, be given conjugate 
diameters of an elliptical shadow, represented by oblique projec- 
tion, as diameters of the shadow of a vertical circle, upon the 
horizontal plane which is represented by the part of the paper 
below AT)' ; a parallel to mn, through Q, as in Fig. 66. Then 
we have om, equal and parallel to the radius of the original circle 
casting the shadow, and Q, as the shadow of the foot of the ver- 
tical diameter of the same circle. Then erect QP, perpendicular 



38 GENERAL PROBLEMS. 

to AT>', and equal to mw, and OQ^iPQ is the radius of the 
original circle. 

Next, lines that are parallel in reality, as ON and a tangent at 
P, are parallel in projection ; and conjugate diameters are, each, 
parallel to the tangents at the extremities of the other. Hence 
any diameters at right angles to each other, in MPNQ, will cast 
shadows which will be conjugate diameters of the given ellipse. 
"We therefore seek a pair of such diameters, whose shadows shall 
be conjugate diameters at right angles to each other, for these 
will be the axes required. 

Now, since the shadows of the indefinite radii of MPNQ begin 
in AT)', these radii, and their rectangular shadows, must be in- 
scribed in two semicircles, having a common diameter in AT)\ 
Hence Oo is a chord of the circle A'OD'o, thus composed, an. I 
gGr, perpendicular to it at its middle point, meets AT)' in the 
centre, Gr, of this circle. Then OA' and OD' are the rectangular 
radii, whose shadows, A'o and D'o, are at right angles to each 
other, and are each parallel to the tangents at the extremities of 
the other. Hence, limiting them by the rays aA and (£D, and 
making oB— oA, and oC=oD, we have AB and CD for the 
required axes of the elliptical shadow mQnp. 

^Remark. — The construction just given is often found among 
plane problems on the conic sections, as it can readily be explained 
by the principles of plane geometry. But, as required by the 
spirit of the present subject, it is here explained by the principles 
of projections and of shadows. 

4A. The problem of the Niche is a favorite one in Shades and 
Shadows, owing to the considerable number of points of peculiar 
interest connected with it. The niche is a familiar concavity in 
the walls of halls, staircases, etc., and consists usually of a verti- 
cal half cylinder of revolution, united by its upper base with a 
concave spherical quadrant. The problem of the niche, as a 
problem of shadows, consists in finding the shadow of the edge 
of shade of the cylindrical part upon that part, and upon the 
lower base ; and the shadow of the front semicircle of the 
spherical part, upon that part, and upon the cylindrical part. 

45. Agreeably to the classification of (12, 13) it becomes de- 
sirable to divide this problem, and to retain here the construction 
of the shadows on the cylindrical part, only leaving the shadow on 
the spherical part to be placed with shadows on oth er double-curv- 



SHADES AND SHADOWS. 39 

ed surfaces. One of the topics of special interest, in this problem, 
is the direct construction of that point of shadow which falls on 
the curve of the upper base ; and there are several such construc- 
tions, but which are applicable only when the spherical part is 
included in the problem. Hence it becomes quite desirable tc 
find also a direct construction of the point in question which 
shall be applicable in the problem as given below, where the 
spherical part is not recognised. Such constructions will be found 
in the following problem. 

Peoblem XVIII. 

Having a vertical semi-cylinder with its meridian plane parallel to 
the vertical plane of projection, it is required to find the shadow 
cast on its base and visible interior, by its edge of shade, and by a 
vertical semicircle, described on the diameter of its upper base. 

Principles. — All the points of shadow, save that on the circum- 
ference of the upper base (45), are found by the special method 
of (40). 

Constructions.— PL VI., Fig. 20. AdB— A'A^B'B" is the 
vertical half cylinder ; AB — A^F'B" is the vertical semicircle, 
whose shadow is to be found, and Aa — A." a" is a ray of light. 

1°. To find the shadows of the vertical line, A — A' A". Aa 
is the horizontal trace of a plane of rays through the vertical line, 
A — A' A". This plane will, therefore, cut the cylinder in an 
element whose horizontal projection is a. Hence, by drawing 
a' — a" , the vertical projection of this line, and the rays, a!V and 
A!' a", we find the whole rectilinear shadow, a' — a", on the cylin- 
drical surface, also the portion A!'b' of the line A' A", which casts 
the shadow a' — a' The shadow of A!b' , on the base of the 
cylinder, is Aa. 

2°. To find points of shadow cast by the semicircle AB — 
A^F'B". Qc — CV is a ray, through any assumed point, CC, of 
this semicircle. It pierces the cylinder at the point whose hori- 
zontal projection is c, and vertical projection c' (40) ; cc' is there- 
fore the required shadow of CO'. Other points of this shadow 
are found in the same manner. 

3°. To find the point of shadow on the upper base of the cylinder, we 
will first use the following special method of auxiliary oblique projec- 
tions. The ray CS — C'S' pierces the plane of the upper base of the 



40 GENERAL PROBLEMS. 

cylinder at S'S. If then the semicircle, AB — A"F'B", be re« 
volved forward about AB — A"B" as an axis, till it coincides with 
the plane of the upper base, CO' will appear at C", and C"S 
will be an oblique projection of the raj CS — CS', when the 
projecting lines come directly forward and downward at an angle 
of 45° with the horizontal plane. Observing the limitation of 
C"S by CS, it may be described as being inscribed in a circle, 
U — SKC", whose centre, U, is found at the intersection of AB 
with iiU, the bisecting perpendicular of CS. Now a parallel ray, 
similarly inscribed in the circle E — BAG", will evidently be the 
similar representation of a ray which contains a point of the semi- 
circle AG /7 B casting the shadow, and of the semicircle AdB 
receiving the shadow. The point on Ac?B will be the point of 
shadow sought. This ray, parallel to C"S, will be a homologous 
side of a triangle similar to SCO", and similarly situated. From 
these things, it follows that the two rays will be to each other as 
the radii, UK and EA, of the circumscribing circles SKC" and 
B AG". Constructing this proportion at qmo, we find mq for the 
length of the parallel ray, dG". Its extremity, d. is located by 
drawing Ec? parallel to US. Projecting d at d', gives dd' as the 
required point of shadow on the circumference of the upper base 
of the cylinder. 

Remark. — To give completeness to the figure, draw dG" parallel 
to SO". It will be equal to qm. Then make the counter-revolu- 
tion of AG"B to its primitive position, when G" will be found in 
its primitive position at GG'. Then drawing the projections, Gd 
and G'd', of the ray which determines dd', we shall have the point 
GG', whose shadow is dd', and, in plan, the complete triangle 
G"Gd, similar to CCS. 

4°. By another special method, that of auxiliary spheres, the 
ray CS — CS' pierces the plane of the upper base of the cylinder 
at S'S, as before. A plane, perpendicular to this ray, at its middle 
point rr', will intersect AB — A"B" in the centre of a sphere of 
which the ray is a chord. Knowing (Des. Geom. 536) that the 
vertical trace of this plane will be perpendicular to CS', the line 
r'T' — rT, where r'T' is perpendicular to CS', and rT parallel to 
the ground line — is a line of the plane, hence T'T is a point of 
its trace on the plane of the upper base. This trace must be per- 
pendicular to CS, hence it is the line TU. ■UU / is then the 
centre of a sphere, whose radius is US, and in which the ray 
CS— CS' is inscribed. 



SHADES AND SHADOWS. 41 

To find the ray similarly inscribed in a sphere whose centre ia 
E,E ; and radius EA. 

In the proportion UK : EA:: CO" : GG" {=gG') we find, by a 
construction like qmo, the fourth term, and locate it as at gQ'. 
Then draw the ray Gd — G'd', which intersects AdB — A"B" a 
<M the point sought. 

Remarks. — a. By finding the shadow on the cylindrical sur- 
face produced, as at// 7 , the intersection of the sketched line of 
shadow, a"c'f , with A^'B", gives 6! approximately and indirectly. 
Such constructions are less satisfactory than determinate inter- 
sections, found by direct constructions, such as the two preced- 
ing. 

b. The straight shadow, a' — a", and the curved one, d'ca'\ are 
tangent to each other at a" '; for a— a' a" is the element of con- 
tact of a vertical tangent plane at a, and the surface of the given 
vertical cylinder, and Ka—h!'a" is the element of tangency of the 
vertical plane of rays, Aa, with the semi-cylinder of rays having 
the semicircle, AB — A^F'B", for its base. Hence a— a' a" is the 
intersection of the tangent planes to the two cylinders, and is 
therefore also the tangent line to their curve of intersection ; viz. 
to the shadow al'c'd'f. (D. G. 170.) 



SECTION II. 

SHADES AND SHADOW'S ON WAEPED SUKFACES. 

§ I. — Shades. 

46. The line of shade on a warped surface will, in general, be 
a curve of some kind. Its points must, therefore, be separately 
found. But any point of a curve of shade may be regarded, 
either as the point of contact of a tangent plane of rays, or as that 
of a single tangent ray. But, again, since we naturally associate 
tangeucies of surfaces to surfaces, and of lines to lines, a tangent 
line to a given surface at any point, is generally constructed as a 
tangent to some curve, lying on that surface and passing through 
the point. 

47. Here it is to be remembered that, as the consecutive ele- 
ments of a warped surface are not in the same plane, a plane, K, 
passed through any one of them, E, will, in general, intersect al) 



42 GENERAL PROBLEMS. 

the others on each side of it, forming a curve, C, whose inter 
section with the element, E, is, as explained in Descr. Geom 
(212, 342), the point of contact of the plane K, which is thus a 
tangent plane as well as a secant plane. 

48. Hence we have two direct general methods (22) for deter- 
mining any point of the curve of shade on a warped surface. 
One of these methods will be given here, the other presently. 

First Method. — Construct the point of tangency of any plane of 
rays, tangent to the warped surface, by passing a plane of rays 
through any element, and noting the point of intersection, P, of thai 
element, with the curve of intersection of the tangent plane with the 
warped surface. The point of intersection, P, will be the point of 
tangency of the plane of rays, and hence a point of the required curve 
of shade. 

49. When the warped surface is of the second order, either a 
warped hyperboloid, or a hyperbolic paraboloid, the plane of rays 
passed through any element, intersects the surface in a second 
element, of the other generation (Des. Geom.), whose intersection 
with the given element is the point of tangency of the given 
plane. 

In illustration of the method of (48) take the following simple 
case. 



Problem XIX. 
To construct the curve of shade on a hyperbolic paraboloid. 

Principles. — By taking the planes of projection as plane direc- 
tors, and one of the directrices vertical, the construction will be 
simple. The horizontal traces of the planes of rays containing 
the horizontal elements will be parallel to those elements, respec- 
tively ; and will cut the horizontal trace of the surface in points 
of the elements which are parallel to the vertical plane. 

Construction. — PI. "VI., Fig. 21. For the sake of definiteness, 
call the horizontal plane the plane director of the first generation, 
and the vertical plane, the plane director of the second generation. 
Let the vertical line A — A" A' and the line A"G — A/G, in the 
vertical plane, be the directrices of the first generation. Then 
divide these directrices proportionally, according to the proper- 
ties of the surface, and, for convenience, let these parts be equal 



SHADES AND SHADOWS. 43 

on each line. Then AA^-A'; AC— o"0'; AD-cTD', etc., 
will be elements of the first generation. In order that the curve 
of shade shall be on the visible face of the surface, or real, k 
case of an opaque solid, formed and placed as in the figure, the 
light should come as shown at A7 in horizontal projection. 
Without constructing the intersection of the raj through AA' 
with the horizontal plane, by means of a given vertical projec- 
tion, let 7 be assumed as this intersection. The line 7w, parallel 
to AA ;/ , will then be the horizontal trace of the plane of rays 
through AA". Then na, drawn from n, the intersection of the 
trace, 7w, with the trace, AG, of the paraboloid, and parallel to the 
ground line, represents the element of the second generation con- 
tained in the plane A — 7n. Its intersection, a, with the element 
AA", is the point of contact of the plane of rajs, and hence a 
point in the required curve of shade. 

Since the vertical directrix at A was equally divided bj the 
horizontal elements, the rajs through the points of division, 
being parallel lines, meet the horizontal plane in points, on A7, 
which would divide that line into equal parts, and which are also 
points in traces of planes of rajs through those elements respec- 
tivelj. Hence 6p, parallel to AB, and le, parallel to AD, are, 
for example, horizontal traces of planes of rajs containing those 
elements. These planes also contain, respectivelj, the elements, 
bp and eo?, of the second generation, which intersect the former 
elements at b and d, two more points of the curve of 
shade. 

Other points of shade are similarly found. The vertical pro- 
jection of b is at b' ; of d 7 at d\ etc. AN is the horizontal pro. 
jection of &'N', the first element below the horizontal plane. 
IK is the horizontal projection, parallel to AG, of the trace of 
the plane of rajs through AG on the horizontal plane h'W. 
Then bj drawing k#, as in the previous constructions, we find g ) 
the point of shade on AG. 

Remarks. — a. From the properties of this surface, the curve of 
shade, that is, geometricallj speaking, the curve of contact of a 
cylinder with a hyperbolic paraboloid, is a parabola, and hence 
is also a parabolic curve in both projections. 

b. The problem, as here given, shows the shadow, as on an 
awning or porch roof in an angle of a building, and of the form 
of a simple hyperbolic paraboloid. 

50. Second Method (48). — Pass any secant plane of rays through 



4:4: GENERAL PROBLEMS. 

the warped surface and construct its curve of intersection xoith th<t 
surface. Then draw a ray tangent to this curve of intersection. 
The point of contact, thus found, will he the point of contact of the ray 
with the surface, and hence a point of its required curve of shade. 

Either of the above direct and general methods (48, 50) can be 
easily applied to any of the general warped surfaces, with or with- 
out plane directors, and given by their elements, in constructions 
which are too simple to need formal illustration. 

51. When the warped surface is a hyperboloid, particularly 
when it is a hyperboloid of revolution, the indirect special method 
of auxiliary tangent surfaces is applicable. This method is indirect, 
as involving an intermediate surface whose line of shade is more 
readily found than that of the given surface (22), and is special, as 
being applicable to certain particular surfaces (22). It may be 
stated as follows : 

Make any circular section, C, of the given warped surface of revo- 
lution, the circle of contact of an auxiliary tangent cone, having the 
same axis as the given surface, and whose elements of shade, E and E', 
are easily found. 

The point, T, at which the element of contact, E, of the auxiliary 
cone with & plane of rays, intersects the circle of contact, C, of the 
cone and the given surface of revolution, is the common point of 
contact of the plane, the cone, and the given surface. That is, it 
is the point of contact of the plane of rays and the given surface, 
and hence is a point of the curve of shade on the latter. 

Hence, the essential principle of the above method, stated 
abstractly, is this : Two surfaces, M and N, being tangent in a curve 
of contact, K, the intersection, T, of the line of shade of N with the 
curve of contact, K, is a point of shade, common both to M and ~N t 
and hence a point of the curve of shade ofM. 

Remark. — The method just explained might be applied in con- 
structing the curve of shade of a warped hyperboloid; but as it 
will be exhibited in connection with an analogous double curved 
surface (Prob. XXVII.) it is here omitted. 



PI .VII 







ri.vn. 




SHADES AND SHADOWS. 45 



Problem XX. 

To find the curve of shade on the common oblique helicoid, in tht 
practical case of the threads of a triangular-threaded screw. 

52. Principles. — Under this head are here arranged a few pre 
liminary matters, which will prepare the way for attention to the 
immediate object of the problem. 

1°. Description of the screw. PL YIL, Fig. 22. — Let the circle, 
whose radius is AG", be the horizontal projection of a vertical 
cylinder, called the cylinder, core, or newel, of the screw. Let 
w >"l'f>rj\"i \y G an y isosceles triangle, whose base, w"'T"\ coincides 
with an element of the cylinder, and which lies in a meridian 
plane of the cylinder. Let this triangle have a compound 
motion ; of rotation about the axis, A — A' A", of the cylinder, 
and of translation parallel to that axis ; and let each of these 
motions be uniform. With such motions, each point of the 
triangle, as w'", V", or T'", will describe a helix (Des. Geom. 308) ; 
the side l" ,r £'", not shown, but corresponding to w'"1!" (dotted), 
will generate an upwardly converging zone of an oblique heli- 
coid, and the side w'"U" (dotted) will generate a downwardly 
converging zone of a similar helicoid. The entire triangle will 
generate the volume called the thread of the screw. The heli- 
coidal zone generated by l //; T ;// is called the upper surface of this 
thread, and that generated by w'"l'" is its lower surface. The 
curve, w' n Wa'r' , etc., generated by w'", is called an inner helix ; 
while the curve, l"'b"l", generated by V" , is called an outer helix. 

All the inner helices are horizontally projected in the circle 
Warn, and all the outer ones in the circle sqx. The vertical pro- 
jection of a helix is found by dividing its horizontal projection 
equally, and projecting the points of division upon the successive 
equidistant horizontal lines, which mark its uniform ascending 
progress. 

The contours of the helicoidal surfaces are curves, apparently 
tangent, as seen in vertical projection, to the helices. That por- 
tion of the contour which bounds a single zone is, however, so 
slightly curved that it is sufficiently accurate to represent it by a 
straight line tangent to an outer and an inner helix, as at U"w'" 
(the full line) and V'Il 1 ". By thus drawing all the visible con- 
tours, the projections of the screw will be completed. The dotted 



46 GENERAL PROBLEMS. 

line, V"w'", whose horizontal projection is AS", is an asymptote 
to the contour of the helicoid, being an element of the cone direc- 
tor (Des. Geom. 333— 4th) whose axis is A— A' A", so that any 
meridian plane cuts elements from it and from the helicoid, 
which are parallel. 

The core of the screw is shown at Y ,f/r E r// for a short distance 
above the horizontal plane Jc"C\ which cuts off the screw. The 
intersection of the screw with this plane, is a spiral of Archi- 
medes (D. G. 3395), and is found by dividing S"S'" into any 
number, as eight, of equal parts, and the semicircle, S'^G^X . 
into the same number of equal parts ; then circles, with A as a 
centre, drawn through the points of S"S'", will intersect the 
radii through the corresponding points of S^G'^ST"'' in points of 
the spiral required. 

2°. To construct any particular element. — First Method. — Any 
line, as Aae, is the horizontal projection of an element (D. G. 339), 
using a, a moment, to mark a point on Ae. Project a at a', and 
e at e\ then ae—a'e' will be the projections of the portion of the 
indefinite element Ae, which lies on the zone bounded by the 
helices through the points S",S"" and S"> . 

Second Method. — Produce the element through &"JS>"" and 
S'",^'" (not shown in vertical projection) till it meets the axis, 
A — A 'A", at a point which we will call 2. Then the element 
through any point, as e,e\ will meet the axis as far above 2, as e 
is vertically above S' /;/ . This follows from the uniformity of the 
two motions of the element (1°). 

Remark. — As the construction of a single point of the curve 
of shade of the screw is somewhat lengthy, by any method, the 
principles of each of the constructions now to be explained, will 
be given in immediate connection with the construction it- 
self. 

53. Methods by assumed helices. These special methtods 
consist in finding the points of contact of planes of rays, which 
are tangent at points on assumed helices. 

The following will suffice for illustration: 

First. The method by tangent planes of given declivity. 

Principles. First. The axis of the screw being vertical, the 
elements, and the tangents, at all points of the same helix, make, 
each, a constant angle with the horizontal plane. Second. Hence 
all the tangent planes to the screw, at points of the same 



SHADES AND SHADOWS. 47 

helix, make a constant angle with the horizontal plane; they 
being determined by the lines just mentioned. Third. Hence, 
again, the angle between the element, and the line of greatest 
declivity, in these tangent planes, is constant. This line of de- 
clivity is perpendicular to the horizontal trace of the plane. 
Fourth. A cone, having the axis of a helicoid for its axis, and 
whose declivity is the same as that of all the tangent planes along 
a given helix, may be taken as a cone director of the helicoid, 
since its elements make a constant angle with those of the heli- 
coid. Any plane, tangent to this cone will be parallel to one of 
these tangent planes to the helicoid. Therefore, & plane of rays 
being made tangent to this cone, a parallel plane of rays may be 
drawn, tangent to the helicoid, and its point of contact will there- 
fore be a required point of shade. 

From these principles we have the following operations : Draw 
any plane, P, tangent to the screw at a point on a given helix, 
and note its declivity, L, taken in a meridian plane, and which will 
give that of the cone director. Then, draw a plane of rays, 
tangent to this cone. Its declivity, I/, will be parallel to that 
of the required tangent plane of rays, and in the same meridian 
plane. Then the element, whose angle with 1/ equals the angle 
between L and the element of contact of P, will intersect the 
helix at the point of contact of the plane of rays, that is, at a 
point of shade. 

Construction. — PI. VII., Fig. 22. 1°. Let the assumed helix be the 
outei* helix, S"eq — S'"V. At any point, o,o', of this helix, draw the 
tangent line, oO, and the element as Ao. The tangent pierces 
the horizontal plane of projection at 0, where oO is equal to the 
arc osS" (Des. Geom. 317), and the element pierces the same plane 
at F,F'; hence FO is the horizontal trace of the tangent plane to 
the upper surface of the thread at o,o'. Now AG, perpendicular 
to FO, is the line of greatest declivity of this tangent plane, and 
GAF represents the angle between this line and the element of 
contact, AF, of the tangent plane. 

Let the cone, whose axis is A — A' A/' and whose declivity is 
equal to that of AG, be the cone director for the helicoidal upper 
surface of a thread. AG meets the axis at the point A,F" ; 
then let AI — A "I', having the same declivity as AG, be the gene- 
ratrix of the cone director. The circle with A as a centre, and 
radius AI, will be the base of this cone. Through its vertex, 
AA", draw the ray AB — A^B', which pierces the horizontal 
plane of projection at B',B, hence BD is the horizontal trace of a 



4S GENERAL PROBLEMS. 

plane of rays, tangent to the cone director on AD, which is also the 
greatest declivity of this plane, and is the horizontal projection of 
the line of greatest declivity of the parallel plane of rays, tangent 
to the screw. Then lav off. on the outer helix, and from AD, an 
arc 15, equal to u'"o, and the point b will be found as the point of 
contact of a plane of rays, tangent to the upper surface of the 
thread, at a point on the outer helix. Hence 5, and its vertical 
projection, b\ are the projections of the required point of the curve 
of shade on the upper surface of the thread and on this helix. 

BC is the horizontal trace of another tangent plane of rays tc 
the cone director of the upper surface of the thread, and CA is 
its element of contact, and line of greatest declivity. Therefore, 
make C""q equal to u'"o, and q.q will be another point of the 
curve of shade of the npper surface of the thread, and on the 
outer helix, 

2°. To find a point of the curve of shade on the upper surface of a 
thread and on the inner helix, S"'Gr"a — w"'Gr'a'. To simplify the 
construction, take an auxiliary horizontal plane, p'J f , as far 
below n.'n. on the element Ao, before nsed, as the horizontal 
plane of projection is below o,o'. Then, drawing AO, the line 
riP will be the horizontal projection of a tangent at njn\ which 
pierces the plane p'J' at P. For, by the properties of the 
helix and its tangent, n? = nGr"S"' and oO = osS", while 
7»G"S'" : asS":: An : Ao 
::AP: AO 

hence n? : oO:: AP : AO, 
a proportion which is evidently constructed by limiting the 
parallel tangents, at n and o, by the straight line AO. 

The element, An 7 , pierces the plane p'J' at p\p ; hence ^P is 
the trace, on the plane ^/J, of the tangent plane to the thread at 
n^' ; and A2, perpendicular to p¥, is the line of greatest declivity 
of this plane, ^ow, as before, BD is the horizontal trace of the 
plane of rays, tangent to the cone director, and AD is the hori- 
zontal projection of its line of declivity, and of that of the re- 
quired parallel plane, tangent to the screw. Then make 4a equal 
to 3rc (53) and a,af will be the required point of the curve of 
shade of the upper surface of a thread and on the inner helix. 
?\r' is a similar point of the other curve of shade, on the back 
part of the same zone. 

Remarks. — a. Points on other helices, such as intermediate 
ones, can be similarly found. 



SHADES AND SHADOWS. 49 

b. By going through with the preceding constructions, as 
applied to the lower surface of the thread, it will be found that, 
in passing from the element of contact to the line of declivity of 
a tangent plane, the motion will be, in a (rotary) direction, around 
A, opposite to the direction ou" r . Hence, DA, the line of decli- 
vity of the tangent plane of rays to the cone director, being pro- 
duced to x, Ax is the horizontal projection of the declivity of this 
plane, and of the parallel plane of rays tangent to the lower sur- 
face of a thread. This being so, we proceed, according to the 
former statement, to make xl equal to lb, and ym equal to 4$ ; 
then Im is a curve of shade on the lower surface of a thread. 
Likewise, hj is the horizontal projection of a second curve of 
shade on the same surface. 

c. Each of the curves of shade, now found in horizontal pro- 
jection, has as many vertical projections, and corresponding 
curves in space, as there are helicoidal zones in the assumed por- 
tion of the screw. Thus, ab is vertically projected at a'b' , and 
a"b' f ; qr, at q'r and in part near g' and s' ; Im, at I'm' and 
11' m" ; and hj, at hj' and in part at Jc"j", and m!" . 

d. Tf we conceive the entire helicoid, of which the upper sur- 
face, as S""tv'"a'e', of any thread is a zone, to be represented, 
we may proceed as before to find other points of its curve of 
shade. The point A,D', where the curve of shade meets the 
axis, is the intersection of the axis with the element through 
which the meridian plane of rays, EAK, is passed, since, as the 
axis and this element are the two lines cut from the surface by a 
plane of rays containing the supposed element, their intersection 
is the point of tangency of the plane of rays (D. G. 342), and is 
therefore a point of the curve of shade. We thus find baAjh — 
b'a'D' (the rest of the vertical projection not shown) for a com- 
plete branch of the curve of shade on the helicoidal surface 
bounded by a helix of a radius equal to AS". Likewise ImArq 
— I'm' is the branch of the curve of shade, on the similar heli- 
coid of which the lower surface of a thread is a zone. 

e. The point, A,D', is also the point of tangency of the plane 
of rays RAK, from the particular properties of the helicoid, as 
well as from those of warped surfaces in general, as explained in 
the last remark. For the axis A — A 7 A" is the limiting helix, 
generated by that point of an element which is in the axis ; but, 
being straight, it is its own tangent, throughout its whole extent; 
hence, A,D', like other points of contact, e,e', etc., of tangent 

4 



50 GENERAL PROBLEMS. 

©lanes, is the intersection of a tangent to the helix, through A,D', 
with the element through the same point. 

54. Second. The method by helical translation of a tangent 
plane. 

Points of shade, on the lower surface of a thread, may be found 
by the method of [53-2° (Rem. b)~\. For the sake, however, of 
illustrating a different construction, they will be found by the 
special method just named. 

Principles. — Construct a simple tangent plane, P, to the 
helicoid at any point of a given helix, find its trace, T, on any 
horizontal plane, K; and note its intersection, T, with the axis of 
the helicoid. A ray through I will pierce the plane K in the 
trace, T, if P is a plane of rays. If this ray does not meet the 
trace T, revolve it about the axis of the helicoid, when it will 
generate a right cone, from which will be cut two elements, E and 
E', by the plane P. By revolving either of these to coincide with 
the ray, and by revolving the point of contact of P through an 
equal angle, keeping it also on the helix, we shall have the point 
of contact of P, when translated, helically, so as to be a tangent 
plane of rays. This point of contact will therefore be a point of 
the required curve of shade on the given helix. 

Construction. — PL VII., Fig. 22. Let the tangent plane be 
constructed at the point S",Z'", and let its trace be found on the 
horizontal plane S'K'. Thus S"A — £'"U, the element, through 
S'V" of the helicoidal zone, forming the lower surface of the 
thread considered, pierces the plane S'K' at S',S. The plane 
S'K' being here taken at a distance above S'T" corresponding to 
a quarter revolution of S",^" around the axis A — A! A." we have 
S // T //// , equal to a quadrant of the circle S"^, as the tangent at 
S'T", piercing the plane S'K' at T"". Hence T""ST is the 
trace, on the plane S'K', of the tangent plane at S'V". Now, 
as this tangent plane contains the element ^'"U, it cuts the axis 
at U, and we wish next to find whether a ray through U pierces 
S'K' in the trace TT"". For lack of room on the plate, draw 
AK — UK', the position of a ray through A,U, after a revolution 
of 180° from its primitive position, AK. (The angle K'UA' is 
equal to the angle B'A"A'.) The revolved ray pierces the plane 
S'K' at K',K, one point of the base, KQT, of the' cone generated 
by the revolved ray AK — U'K. As this base does not contain 
R, the ray in its primitive position, AR, does not pierce the 



SHADES AND SHADOWS. 51 

plane S'K', in the trace, TT //7/ , of the tangent plane. Hence this 
tangent plane is not a plane of rays. But AT and AQ are 
elements cut from the cone, just described, by the tangent plane, 
and may be considered as two positions of the ray AR, after 
being revolved about the axis, A — A' A", till it becomes a line 
of the tangent plane. Therefore, the ray being at TA, when the 
point of contact of the tangent plane is at S'T", make S /7 A=7— 5, 
and project h at A', then will hji' be the point of contact of the 
tangent plane after being revolved, with an ascending helical 
motion about A — A 7 A", till by the return of TA to being a ray 
AR, it becomes a tangent plane of rays. Likewise, AQ being 
the revolved position of the ray, RA, when S",£ //v is the point of 
contact of the tangent plane, make S"7=8 — 5, and project I at l\ 
and IJJ will be another point of contact of the tangent plane, 
after moving, with a descending helical motion, till it becomes, 
again, a tangent plane of rays. Hence, finally, h,h' and l,V are 
points of the two curves of shade on the lower surface of a thread 
of the screw. 

In a similar manner j,j' and m,m! could have been found. 
hj — h'j\ being in front of the meridian plane SI, is visible in 
vertical projection. Ira— V ml is wholly invisible. 

55. A little consideration of the properties of the helicoid, 
explained in Descriptive Geometry (Des. Geom. 339), will enable 
the student to understand the following additional properties, 
which are here applied in another construction of the required 
curve of shade, by the special method^ called the method hy 
assumed elements. 

Principles. — The development of a given portion of any helix 
— its axis being conceived to be vertical — is the hypothenuse of 
a right-angled triangle, whose altitude is the vertical height 
between the extremities of the given portion, and whose base is 
the development of the circular arc forming the horizontal pro- 
jection of the same portion. 

56. Portions of different helices on the same helicoid, but 
included between the same meridian planes^m^j be called homo- 
logous. Points in which they intersect the same element, may be 
called homologous, or corresponding points. 

Since all points of the generatrix of a helicoid ascend at the 
same rate, the homologous portions of helices just mentioned, 
will be of equal height, between their extremities ; that is, they 



52 GENERAL PROBLEMS. 

will be included between pairs of equidistant horizontal planes. 
Tangents to these helices at corresponding points, and limited by 
the horizontal planes which include these helices, may be called 
homologous tangents. Their horizontal projections will be par- 
allel, and equal to the concentric circular arcs which are the 
horizontal projections of the respective helical arcs to which they 
are tangent. But these arcs, and hence their tangents, are pro- 
portional to their radii ; hence the horizontal projections of these 
tangents will terminate in a straight line through the horizontal 
projection of the axis (53-2°), which, with the tangents, and 
their respective radii, will form similar triangles. 

06. Once more, equidistant horizontal planes will intercept 
equal portions of any element ; hence, if two such planes contain, 
respectively, the point of contact and the trace of a tangent 
plane through any element ; and if another pair of such planes 
contain the point of contact and the trace of another tangent 
plane through the same element, the distance between the trace 
and the point of contact of one tangent plane will be equal to the 
distance between the point of contact and the trace of the other 
tangent plane, these distances being measured on the element 
contained in both of these planes. Eecollecting, too, that the 
helix coincides with its tangent when developed, we have the 
following. 

Construction. — PL VII., Fig. 22. Assume any element, Ae, 
between the previously found points, a and b, on the helices. 
Making the tangent, <?L, equal to the arc esS" ; the point L 
is the intersection of this tangent with the horizontal plane of 
projection (D. G. 317). Then H'H being the intersection of the 
element Ae — EV (52-2°) with the horizontal plane, HL, is the 
horizontal trace of the tangent plane to the screw, at ee' (D.Gr. 347.) 

Next, by drawing a ray (not shown in vertical projection) 
through A,E', it will pierce the horizontal plane of projection at 
6 in the line AB, and 6HM will be the horizontal trace of a 
plane of rays through the assumed element. 

Now, if the trace, HM, of this plane of rays, were found on a 
horizontal plane as far below its required point of contact, as the 
trace, HL, is now below ee\ this new trace would be parallel to 
HM, the tangent at its point of contact would be parallel to eM., 
and their distance apart, measured on Ae, would be equal to eK. 
Hence the portion of the new tangent, cut off by the new trace, 
would be equal to 0M. But all tangents to helices at points on 



SHADES AND SHADOWS. 53 

Ae, and included between pairs of equidistant horizontal planes, 
will terminate in AL, as seen in horizontal projection (53-2°) ; 
hence, transfer M to N, on AL, by a line MIST parallel to HA, 
draw Nc, parallel and equal to «M, and c,c f will be the point of 
contact of the plane of rays, HM, with the screw, and on the 
assumed element Ae. Hence c,c ! is a point of the required curve 
of shade. 

JRemarh — By the above method, as well as by that of (53-2°), 
we can find the indefinite curve of shade on a complete helicoid. 

Discussion. 



First. The meridian plane, EAK, which is tangent to the heli- 
coid at AjD 7 , makes an angle of 90° with the horizontal plane. 
At a point, analogous to any point as e. but on the helix which 
is at an infinite distance from the axis, the tangent eL would be 
infinite, hence HL would then be parallel to eL ; that is, AH 
would be both the line of declivity and the element containing 
the point of contact of the tangent plane. Hence when the 
element of contact of a plane, tangent to a helicoid, is also the 
line of declivity of that plane, the point of contact is on a helix 
at an infinite distance from the axis. 

Second. To find the helix, H, the tangent planes for all points of 
which make a given angle, /3, with the horizontal plosne ; /3 being 
greater than oc. Construct a cone, having the axis A — A' A" for 
its axis, and its vertex at the intersection of an element, E, of the 
helicoid, with the axis, and let all its elements make an angle 
equal to /3 with the horizontal plane of projection. Then con- 
struct a plane, through the element E, and tangent to this cone. 
This plane will make an angle /3 with the horizontal plane, and 
its point of contact, on E, found as in (57 Const.), will be a point, 
p, of the required helix. Having the point p, the helix can 
readily be constructed. 

Third. Now let /3 be the angle made by the rays of light with 
the horizontal plane of projection, and ex, as before, the angle 
made by the elements of the helicoid with the same plane. The 
angle /3 may be greater than, equal to, or less than oc. In J?]. 
VIL, Fig. 22, /3 is less than oc, and the curves of shade are open 
curves with infinite branches. If j3 were equal to oc, a tan- 
gent plane of rays would be tangent to the helicoid at the inter- 



54 GENERAL PROBLEMS. 

section of the element, AZ", with the infinitely remote helices, 
one on each nappe of the helicoid. At the lower of these points, 
b and q would unite ; and at the upper one, I and h would unite. 
When /3 is greater than oc, find the helix, H, described above, 
and apply to it the method of (53-2°), but giving to the elements 
of the auxiliary cone an angle of declivity equal to /3. It will 
then be found, that but one plane of rays can be drawn, tangent 
on the helix H, and that the points q and b will unite on that 
helix. Also I and h will similarly unite on a similar helix of 
the upper nappe, and the curves of shade will be closed curves, 
wholly within the helix H. 

Fourth. If we substitute for the triangular generatrix of the 
thread of the screw, a square, having one of its sides in an ele- 
ment of the cylindrical core of the screw (52), a square-threaded 
screw will be formed, the upper and lower surfaces of whose 
threads — the axis being vertical — will be right helicoids, having 
the horizontal plane for their plane director. For all practical 
cases the angle fi is not 0°, hence for the case of the curve of 
shade on the square threaded screw, /3 is greater than oc, and 
the curve of shade would therefore be closed. Usually, also, /3, 
and the rate of ascension of the threads, have such values that 
the curve of shade would be wholly within the core; hence, 
practically, there is no problem of the curve of shade on the heli- 
coidal surfaces of the square threaded screw. 

The elements of shade on its cylindrical surfaces are readily 
found. 

58. Either of the direct general methods (48, 50) may be applied 
in the construction of curves of shade on warped surfaces, but 
owing to the ease with which a plane is drawn tangent to a 
warped surface of the second order, the following indirect general 
method will be more convenient. 

Observing that a hyperbolic paraboloid may be more easily 
represented in any position than a warped hyperboloid, make 
any element of the given warped surface the element of contact of an 
auxiliary tangent hyperbolic paraboloid* The point of contact of a 
tangent plane through this element, can easily be found, and it will on 
the point of contact on both surfaces. (Des. Geom. 343.) 

In illustration of this method the following problem is given. 



SHADES AND SHADOWS 55 



Problem XXI. 

To construct points of the curve of shade of a conoid. 

Principles. — -Let the curved directrix of the conoid lie in the 
horizontal plane, let its rectilinear directrix be perpendiculai to 
the vertical plane of projection, and let this plane be taken as the 
plane director of the conoid, and of the elements of the first 
generation of the auxiliary paraboloid. Let the directrices of the 
first generation of the paraboloid be the line of striction of the 
conoid, and a tangent to its base at the foot of the assumed ele- 
ment of the conoid, through which a plane of rays shall be drawn. 
The horizontal plane of projection will then be the plane director 
of the elements of the second generation of the paraboloid. 

Construction. — PI. YIIL, Fig. 24. MEK is the curved, and 
A A" — A' is the rectilinear directrix of the conoid. Let TE — T'E' 
be the element of the conoid, on which it is proposed to find a 
point of the curve of shade. Through any point, as A", A', of this 
element, draw the ray A /7 R — A'R'. This ray pierces the hori- 
zontal plane of projection at R, and the assumed element pierces 
it at E,E', hence ERG is the horizontal trace of the plane of rays 
through TE— T'E'. Gh', its vertical trace, is parallel to E'T', since 
that element is parallel to the vertical plane of projection. 

Now, making AF tangent to MEK at E, it follows that 
AA" — A' and AF are the directrices of the elements of the first 
generation of the auxiliary paraboloid, which is tangent to the 
conoid along the element ET — E'T'. 

FA — FATi' is another element of the paraboloid, and it is cut by 
the plane of rays, through ET — E'T', at h'h, which is, therefore, 
a point in an element of the second generation. But the hori- 
zontal plane of projection is the plane director of such elements, 
AF being one of them ; hence h'T' is the vertical projection of 
the element, and T' the vertical projection of its intersection with 
ET— E'T'. Projecting T' at T, we have T,T' as the intersection 
of elements, one of each generation, of the paraboloid. These ele- 
ments, being contained in the same plane, viz. the plane of rays, 
T,T' is the point of tangency of this plane of rays with the para- 
boloid. But, as the paraboloid and conoid are tangent to each 
other throughout the element ET — E'T', the point T,T' is the 



56 GENERAL PROBLEMS. 

point of tangency of the plane of rays with the conoid, also ; and 
hence is a point of the required curve of shade of the conoid. 

Remarks. — a. If the direction of the vertical trace, G'h\ of the 
plane of rays, were unknown, it would be found by drawing, 
through any point of ET — E'T', a horizontal line, as A!'g — Ay, 
parallel to EGr, whose intersection, g,g' , with the vertical plane, 
would be a point of G'A'. 

b. The line hT, being a horizontal line of the plane of rays, is 
parallel to GE, the horizontal trace of that plane. Also, a vertical 
line at A, being a directrix of the second generation of the parabo- 
loid, AT necessarily meets FE at A. Hence to find T,T' we have, 
practically, the following very simple construction. First, draw 
EH. Second, through A draw Ah parallel to Eft. Third, note 
T, the intersection of Ah and ET, and project it at T 7 on E'T". 

Any other points of the curve of shade may be similarly 
found. 

c. It follows, from the constructions now given, that either 
projection of tbe curve of shade of a conoid may be found inde- 
pendently of the other. 

Discussion. 

First. — Preliminary. This discussion may be conducted with 
reference to the right conoid with a circular base, PL YIIL, Fig. 
25, having the vertical plane of projection for its plane director, 
and the line, A A" — A 7 , for its line of striction. 

Let cR be the ray through the point c,A' of the element ec. 
Then eR is the trace of the plane of rays through ec. By 
(Rem. b) above, note a, where the tangent at e meets A" A'. 
Then at, narallel to eR, meets ec at t, the point of shade in the 
branch Ath, and determined by the tangent plane of rays eR. 

The vertical projection of t will be on the vertical projection 
of ec, and in the upper nappe of the conoid. By similar con- 
structions, four branches of the curve of shade may be found, 
whose horizontal projections are Ath] AFra; A"k, and A!'g. 

Second. In proceeding with the discussion, the following casea 
will appear. 

First. The horizontal trace of a plane of rays through AA /j 
— A 7 , may intersect the base of the conoid. 

1st. In any manner, as at EE" PL YIIL, Fig. 25. 
2d. In the diameter A A" — W. 



SHADES AND SHADOWS. 57 

Second. This trace may be tangent to the base of the conoid. 
Third. The same trace may be wholly exterior to the same 
base. 

1st. At a finite distance. 
2c?. At an infinite distance. 

Third. For the 1st form of the first case, it appears: 1°. That 
the points of shade, consecutive with A, A' and A", A', being 
points of tangency of parallel rays, on elements consecuv:ve 
with the vertical elements at A and A"; A, A' and A", A' 
are cusp points of the first species, as seen in horizontal projec- 
tion. 

2°. That the element which pierces the horizontal plane at K,K' 
is, in space, an asymptote to the branches Ah, on the upper, and 
A ,r g, on the lower nappe of the conoid. For, the tangent at K, 
analogous to ea, will be parallel to A"A, and hence the points a' 
and a", analogous to a, and determined by this tangent, will be 
at infinite distances, on A A" in each, direction from J. But 
lines analogous to at, and through such points, a' ,a" , and parallel 
to the trace of the plane of rays through JK — A'K', would meet 
HK — H/'K' only at infinitely distant points of shade, analogous 
to t. Likewise, the element which pierces the horizontal plane 
at H,W is an asymptote to Am and A"h. 

3°. The points AjTjT" and A" are all vertically projected at 
A', hence the branches Ah and A"g, each of which is partly on 
both nappes of the conoid, form loops, as seen in vertical pro- 
jection ; to which N'T' and E'B/ are tangent, at the multiple 
point T'. 

4°. All those parts of these curves of shade are real, or would 
exist on the exterior of a conoidal opaque solid, like the figure, 
from whose points rays would pierce the horizontal plane of 
projection, outside of the base of the conoid. 

For the 2d form of the first case, AA" and A'W would 
be the projections of a ray of light ; and the loops in the vertical 
projection would disappear. 

Fourth. In the second case, the construction of (Fig. 24 — ■ 
Bern, b) will give a curve of one branch, in which T and T" will 
unite at J, and loops will reappear in vertical projection, having 
A'N' and H"K' for their tangents at the point J,T'. 

In this case, the element JK — A'K' will also be an element 
of shade, being an element of tangency of a plane of rays, through- 
out its whole extent. On this element, therefore, the conoid is 



58 GENERAL PROBLEMS. 

said to be a developable single curved surface. Along the ver« 
tical elements at A and A" planes may also be tangent. 

Remark. — Fifth. In the 1st form of the third case, the loops of 
the vertical projection will again disappear, and two branches, 
as seen in horizontal projection, will proceed from A to the right, 
and two from A." to the left. 

In the 2d form of this case, the ray of light will be horizontal. 
The vertical projection will be of the same form as in the lsl 
form of this case, but in horizontal projection, the two branches 
from A, and the two from A", will coincide. 



§ n. — Shadows. 

59. The fundamental condition for determining a plane, is, 
that it shall pass through three points taken at pleasure. Equi- 
valent derived conditions are, that it shall contain a point and a 
straight line; or shall contain one straight line, and shall be paral- 
lel to another. All the elements of a cone have a common point, 
the vertex, hence, as already seen (41 b), it is easy to pass a sys- 
tem of planes ; each containing a given line, as a ray, and the 
vertex; and these will be planes of rays, cutting the cone in rec- 
tilinear elements. Likewise, all the elements of a cylinder have 
a common direction : hence, as before seen (41 a), a system of 
planes of rays can be readily passed through the various elements 
of a cylinder. But let a plane be passed through a given ray, 
and a given point of a warped surface ; only one element would 
generally pass through that point, and that, generally, would not 
happen to be in the supposed plane. Again, let a plane be passed 
through a given ray and parallel to some given line ; only one 
element would generally be parallel to that line, and that element 
would not commonly happen to be in the plane employed. 
Hence we have the Theorem, that, as the elements of a warped sur- 
face have neither a common point, nor a common direction, no simply 
arranged system of planes of rays can be made to cut it, each in a 
right-line element. 

60. Hence (41) in its application to warped surfaces gives the 
following general method. 

Pass any secant plane of rays through any point casting a 
shadow, and, by constructing its intersections with a number of 
elements, find its curve of intersection with the warped surface 



SHADES AND SHADOWS. 59 

Then, the intersection of a ray through the given point, with the 
curve found, will be a required point of shadow. 

61. So much construction for each point of shadow, would be 
extremely tedious, hence we have the following inverse general 
method. Pass a plane of rays through any assumed element of 
the warped surface, and note the point, P, cut by it, from the 
line casting the shadow ; then, the ray through this point will 
pierce the assumed element in a point of shadow. 

It is evident, however, that, unless the plane of rays happens 
to be also a projecting plane, the construction of P will generally 
be as tedious as that of the curve of intersection, in (60). Hence 
various special methods, soon to be explained, are mostly em 
ployed in finding shadows on warped surfaces. 

62. Taking up the topics mentioned in shades on warped sur- 
faces, and in the same order, we have, first, the shadow of the upper 
base of a warped hyperboloid upon that surface, the construction of 
which will be explained in a subsequent problem of double- 
curved surfaces. 

To find the highest and lowest points. Find the intersection of a 
ray, drawn through the intersection of the meridian plane of rays 
with the circumference of the upper base, with the meridian curve 
contained in that plane. 

This construction involves an easy application of the general 
method of (60) whenever the warped hyperboloid is one of revolu- 
tion, with its axis perpendicular to either plane of projection, for 
then its meridian curve is readily found. 

63. To find intermediate points. Employ the special method of 
auxiliary shadows (42). Any horizontal plane, P, will intersect 
the given surface, its axis being vertical, in an ellipse, or circle, 
C, if the surface be one of revolution. The shadow of the upper 
base on the plane P will be a curve equal to that base (33), and 
the intersections of this shadow with C, will be points of shadow 
falling on the given surface. 

64. In PL VI., Fig. 21, the shadow of the curve of shade of the 
hyperbolic paraboloid upon the horizontal plane, is the parabolic 
curve ~Lg, formed by the intersection of the traces of the succes- 
sive tangent planes 7n, etc. L7, equal and parallel to aA, is the 
sharlow of aA, and 7A is the shadow of A — A" A'. 



60 GENERAL PROBLEMS. 

Problem XXII. 

To find the several shadows cast by a triangular-threaded screu 

upon itself. 

65. These shadows are: 
First. The shadow of the nut. 

1st. Of any unknown point on an edge of shade of the nut. 
2d. Of some assumed point on such an edge. 
Second. The shadow of the curves of shade. 
1st. On an assumed element. 
2d. On an assumed helix. 
3d On another branch of the curve of shade. 
Third. The shadow of the outer helix. 
1st. On any assumed element. 
2d. On any helix. 
3d. On any particular element. 
Fourth. There may also be associated with the preceding, the 
shadow of the entire screw and nut, on the horizontal plane of 
projection. 

1st. The determination, by inspection, of all the lines casting 

this shadow. 
2d. The construction of the shadow of any assumed point, 
which will cast a shadow on the horizontal plane. 
Each of the above topics, including both its principles and 
construction, so far as these need be separately mentioned, will 
now be taken up in the above order. 

66. To find the shadow of the nut upon the screw. 
1st. The shadow of any unknown point of the nut. 
Principles. — This shadow is found on the principle, that any line 

L in a plane of rays, E, through an edge, E, of the nut, will meet 
the intersection of R with a given surface, in a definite point of 
shadow of some point in the edge, E ; though the line be not, 
itself, a ray. 

This, which is an inverse method (61) may be called the method 
by assumed lines in planes of rays. It is usually applicable only 
when the line E is straight, since a plane of rays can usually be 
passed through a straight line only (6). 

Construction. — PI. YIIL, Fig. 2 j. The edges of the nut, which 
cast shadows, are the lower ones, projected at JK and KL, and 
portions of JI and LN", all of which are vertically projected in 



SHADES AND SHADOWS. 61 

the line J'L'. Beginning with IJ, this edge, produced, pierces 
the meridian plane of rays OAB at 0,0'. Let OA — O'Q be the 
direction of the light, then Q, the point where the ray OA — O'Q' 
meets the axis A — A' A", is the intersection of this axis with a 
plane of rays through IJ. Hence, any meridian plane of the 
screw will cut a straight element from its surface, and a line from 
the plane of rays. This line will pass through Q, and will inter- 
sect the element in the shadow of some point of IJ. Thus, the 
meridian plane, Am, cuts from the upper thread the line hj — k'f, 
and from the plane of rays, the line raA — m'Q, which intersects 
the former line at V ,1, a point of the required shadow. 

In like manner, the edge JK meets the meridian plane of rays, 
OAB, at a, a'. Then .through a,a' draw a ray, and Q" will be 
determined as the intersection of a plane of rays, through JK, 
with the axis, A — A' A". The meridian plane, AB", now cuts, 
from the edge JK, the point, B'^B'", and from the plane of rays 
the line B"A— B^Q". This line meets the element ED— 
E'T)"', cut from the upper thread, at n'" ,n", the shadow of some 
point of JL. 

Remarks. — a. The points, as e n ,e'", where the shadow leaves 
the screw, at its intersection with the helices, are all found by 
the indirect special method, called the method of intersecting sha- 
dows ; and which consists of the following easy operations. Con- 
struct the shadow of the line, L, casting the shadow — in this case the 
edge of the nut — upon any plane below. Find also, on the same 
plane, the shadow of M, the line containing the required point of 
shadow — in this case the outer helix. Then the intersection of 
the shadow of L with the shadow of M will be a point, through 
which a ray will meet M in a point, m, and L, in a point, I; and 
m will be the shadow of I. The point t,t' is thus found. 

b. This construction is not fully shown at B,A", and e"e' r '. 
By drawing the rays through these points, we find the points 
Y,Y' and n,n' of which they are the shadows. Similarly, the 
points which cast any ascertained point of shadow, may be found. 

c. From Y,Y' the shadow of the nut falls on the next thread 
below, beginning at q,q\ the intersection of the ray through Y,Y' 
with the shadow of the outer helix. 

d. The point e,e' is the shadow of a,a!, and is the intersection 
of the ray ae — a f e f with the element be — b"c" . This one point is 
found by (41) as a meridian plane of rays cuts a straight element 
from a helicoid. 



62 GENERAL PROBLEMS. 

2d. To find the shadow of an assumed point of the nut. 

Let the point be an angle of the nut, as JJ'. Constructions 
for the shadow of such a point prove, on examination, to be 
quite numerous. The following are some of them. 

First By the most simple direct method. Pass a vertical plane 
of rays through J, J 7 . Its intersection with the surface of the 
thread, will be a curve, easily formed by joining the points in 
which the plane intersects several assumed elements of the screw, 
taken, for convenience, in the supposed vicinity of the desired 
point of shadow. A ray through J,J' will meet this curve (not 
shown) in the required point of shadow W,W. 

Second. By an obvious indirect method. Find the shadows of 
points, which may be known to be on IJ and KJ produced. The 
indefinite shadows of these edges, found as in (66 — 1st), will inter- 
sect at a point, W,W 7 , which will be the shadow of J, J'. 

Third. By the following indirect method, interesting in theory, 
but tedious in application. Make J, J' the vertex of a cone, 
generated by the revolution of the ray through JJ', around a 
vertical axis at JJ'. The intersection of this cone with the screw, 
is easily found by joining the points of intersection of several 
neighboring elements of the screw with the surface of the cone. 
The intersection of the ray element with this curve, is its inter- 
section with the screw, and therefore is the shadow of J, J'. 

Fourth. Let a plane of rays be passed through J.J', so as to 
cut a rectilinear element from the screw. That is, as such a 
plane contains a ray through J, J', it is really required to pass a 
plane through a given line, and so as to cut a rectilinear element 
from the screw. The given ray pierces the horizontal plane, 
2i"w' ', at Y. Any line through Y may be considered as the 
horizontal trace — on 7a'" w' — of a plane of rays through J.J'. 
The upper surface of the upper thread, produced, intersects the 
plane, Z'"w r , in the spiral wxyz, and a line from A to any point 
of this spiral is the horizontal projection of an element of this 
thread. 

Next, make J,J' the vertex of a cone, generated by the revo- 
lution, about a vertical line at J, J 7 , of a line, J*, which makes an 
angle with the plane Z"W, equal to the angle made by the ele- 
ments of the screw with that plane. The circle, with radius J^ 
and centre J, is the base of this cone, in the plane 7i'"w' . If 
now a trace, YS, can be drawn, such that SJ and 5A shall be 
parallel, then these lines, bemg the projections of lines of equal 



SHADES AND SHADOWS. 63 

declivity, will be parallel in space. Further, as their tracts, S, 
and 5, are in the trace of the plane of rays, the lines themselves 
are in that plane, and as 5 is also a point in the trace of the 
screw, 5A is the element of the screw which is contained in a 
plane of rays through J,e~P. 

Now, to find YS, draw a number of trial traces V6, Y7, etc., 
and note the corresponding radius vectors, 6A, 7A, etc. Then 
draw J8, J4, etc., parallel to 7A, 6A, etc., and note their intersec- 
tions, 8, 4, etc., with the traces Y7, V6, etc. Then the curve, 
8, 4, etc., is the locus, or place, of all intersections of radii through 
J, parallel to radius vectors through A, with the traces through 
Y. Hence S, where this trial curve, or locus, intersects the base, 
US, of the cone, is the point from which can be drawn the 
required trace, SY, of a plane of rays through J, J', and cutting a 
rectilinear element, 5A, from the screw. Then W,W, where the 
ray, JW — J'W, meets this element, is the required shadow of 
J, J' on the screw. 

Remark. — It is now apparent, from the preceding construction, 
that it reduces itself to this problem of two dimensions. Through 
a given point (Y) in the plane containing a spiral of Archimedes 
and a circle, it is required to draw a line (YS) intersecting the 
spiral and the circle at points, from which parallels can be drawn, 
respectively, to the pole of the spiral, and the centre of the circle. 

67. To find the shadow of the curve of shade on the thread below. 

1st. On any assumed element. 

Principles. — This shadow is found by the indirect special method 
of (Q6 — 1st. Eem. a). Thus, the shadow of the curve of shade 
upon the horizontal plane, will intersect the shadow, on the same 
plane, of an assumed element, supposed to contain a point of 
shadow, in a point, through which, if a ray be drawn, it will 
intersect both the element and the curve of shade. The former 
intersection will be the shadow of the latter. 

Construction. — PI. YIL, Fig. 22. To find that point of the 
shadow of the curve of shade, hf — h'f, which falls on the element 
G"s — GV. Drawing rays, G-"X and sY, not shown in the verti- 
cal projection, through Gr^G', and s,s', the line YX will be found, 
as the shadow of the element upon the horizontal plane of pro- 
jection. Similarly, JcZ is found, for the shadow of hf — hf on 
the same plane. These shadows intersect at t. Then tu is the 
ray which intersects G^s— GV at u,u\ which is the required point 
of shadow. 



64 GENERAL PROBLEMS. 

Bemarh. — By producing the ray, that point of hj—h'f, -whose 

shadow is uu\ could be found. 

2c?. On an assumed helix. This shadow is found in the same 
way as was the one just explained. Thus: let the outer helix, 
soe — s'o'e', be the assumed helix. YZ is a portion of its indefinite 
shadow, found by connecting the shadows of several of the points 
of the helix, on the horizontal plane. This shadow intersects 
the shadow, &Z, of the curve of shade, hj—h'f, at Z. Then the 
ray Zv — v't' meets the helix at v,v' which is the shadow of t' 
(between h and i, in horizontal projection) on the curve of shade 
— produced in the present case. 

3d. On another hranch of the curve of shade. Find, by (66 — 1st.) 
a point, y", of the shadow of the curve of shade hj — h'f produced 
upon an indefinite element, as Ae, beyond b on the supposed 
branch, ah, of the curve of shade. Join this point with the pre- 
viously found points of shadow, juy", of hj — h'j upon the screw. 
The indefinite curve of shadow thus found, will intersect ab in 
the required point, d,d'. 

Bemarh. — To find how far this curve of shadow of hj — h'f is 
real, consider that it begins at/,/, the intersection of hj — h'j' with 
the inner helix /GrV, and is limited by the ray through h,h', 
which meets the shadow, juy"—j'u'v', at p"p'" . Beyond p"p"\ a 
real shadow is cast, by the outer helix through hh'. 

68. To find the shadow of the outer helix on the upper surface of 
the thread below. 

1st. On any assumed element. First Method. — This is the method 
of (66— Is*. Eem. a) and of (67— 1st). It is shown, in PI. Till., 
Fig. 23, in the construction of the point fif. Here AM-C'M' 
is any assumed element produced, on which a point of the pro- 
posed shadow may fall. MP is its shadow on the horizontal 
plane, and R"F is the shadow of a portion of that outer helix, 
whose vertical projection is b'"s" . Then Wfi the horizontal 
projection of a ray through the intersection of these shadows, 
gives/ which, projected in CM', at/ 7 , gives fif as the point of 
shadow required. 

Second Method. — This is a special method which may be called 
the method by helical translation of a plane of rays containing an 
assumed element. 

Principles. — When a plane is perpendicular to either plane of 
projection, all lines in it will be projected, on that plane to which 
the given plane is perpendicular, in the trace of the given plane. 



PL YHI 




VI. vin 




SHADES AND SHADOWS. 65 

Also, the intersection of the plane of rays through the element, 
with the given helix, is the point casting a shadow on that ele- 
ment, at its intersection with a ray through that point. 

Construction.— In PL VIIL, Fig. 23, let AT— I // T / (53—2°) be 
any assumed element. It pierces the horizontal plane at N',N. 
The ray AB — F'B', through the upper extremity of the element, 
pierces the horizontal plane at B',B. Then NB is the horizontal 
trace of a plane of rays through the element AN" — F'N'. Let 
this plane now be revolved till it becomes perpendicular to the 
vertical plane of projection. Its line of declivity, AH, will then 
be horizontally projected in AH". Producing AH, the arc CU 
corresponds to HH". All points of the plane of rays revolve 
equally, therefore make T3 = CU, and project 3 at 3' on the outer 
helix. All points of the plane also rise equally, and at the rate 
of the rise of the helix, in order that it may still continue to con- 
tain an element of the screw. Therefore make Y'Y" equal to the 
height of 3' above T', and draw I /// 3 / H /// . As the plane, NB, is 
now perpendicular to the vertical plane of projection, I /// H /// is 
its trace, and it cuts the outer helix at v!"u n . In the counter 
revolution of this plane, u" ! u" traverses a helical arc whose hori- 
zontal projection is equal to CU, hence make ?// / A = CU, project 
h at h\ and then draw the ray hg — h'g ; , which intersects the 
assumed element AT — A"T' at g r g\ the required point of 
shadow. 

2 c?. On any helix. 

This shadow is found by the method of intersecting shadows 
(67 — 1st). Thus, let the shadow of a spire of the outer helix be 
found on the spire below, of the same helix. F is the intersec- 
tion of the shadows, cast on the horizontal plane, by fragments 
of the outer helixes, near s' and u' '. Then the ray Fs gives s, 
whose vertical projection is s r . The same ray produced — not 
shown in horizontal projection — gives u\u. Therefore s,s f on 
the outer helix, is the shadow of u,u', a point on the spire 
above. In the same way d'\6!" is found. 

3c?. On any particular element. 

The point d,d ', in the meridian plane OA, is found by the 
ordinary direct method, as the intersection of a ray, bd — b"'d\ with 
the element be — b'ti ' . 2,2' may be found either asj^/' was, or as 
g,g f was. sdd" — s'd'^'d'" is now the complete shadow of the 
outer helix, througli b,b'", on the lower thread. The visible 
portion of an equal shadow, similarly situated, is shown on the 

5 



-:-: 



jzr s7A:~ i= =7: — z. :z z ~::\ : . izz^zz z ::i zz: — ztt tz 
:z : Tr i. 

:7 7~z 7:7 . ;-' zz :zz. : A z : : ; :z A~:zz ;,; A.z;,f„ 

- - : ". - ". : : -'-- "..::.;' :~ :/.-: . z z z : : . : zz : z : . 

7_zi lines zz, s: z. .:7 :: z :z :f :- ; zz~z: :: s7: A is is 
rz:7. :::: ;5. ::: ;::::l:i r : s7: : z ~ ;■ z :'. :z ::z zxzrA: :: 
::: _:..;:;:; 5 : z .: :_ ::' fif: : zz" __t 7z z : z\ A: :z 7z 
zzAarzi r zr: A At zzz: 5 :z: z :f ;. :z:z z. rzzit :z:7:.z:z 
i:zi :;z :f :ir 1:— fr zzz: z :'_r zzt 7_:z :. . izi zz, 
:i:>sz ::z z slAt :f z'lz :z: ~ z:A :z ::: :-zs: zz.:~; z: 
the sere - . 

A7 A z zz. z :z zzz:: z' zz _:z z A zz zAz Azzf p;-- 
tio?-^ of zfte M&Z firae of shade of ike screw. 

7_:s zz:::: zz 1 z zz t ; ~ :_t :lz::z zz_ ::.. A 7: z : . HZ 



?.-.:zzzv AA7A. 
.7: :zz"z z. zzz: : z zz zz :~A. : : :.:z" AA: z zz. 

77. 7i ::z:z::z :7f . : 7:z:. : ;-' zz z : 7 ?1. V7A . 77z 75. 
z z z 7t :z::z:t:z z zz. 1= :7 :z TAzizirs z :7t :::::: 
zz:z:: :7t 7:.t :f zzzz:: AA — A 7 t z .7 z ■:- : 

:i:z: t _ :zis TAt izf :z~ :~~ :z:77 zzzzzz z :Azzz: z:z 
the meridian plane, HK. AM such planes will have horizontal 
:: : :z : z 7z. z A A : :. A:.zz :zzz :_t z z :f At :::::: 
A :zz:: :.:.z lAzg z_z :zz: .77 z :_t tz::zzz zzt z 
:_t Azz zz z: A :".i:.-r» 

7::z :7f rrzzzAzg z :A7: — s :zi: :z~ :':z A :z:. :z.zA. 
through the line of stziction, and whose trace, parallel to A - 
z bri— -e-t! AA zA 77 1= :.: 17 ~ A ::::::::: 7_f :::::: z_ :~: 
rectilinear elements. Of these two elements, the foremost one, 
as ET, will be the indefinite shadow of the portion, TA, of the 
line of stziction upon the exterior of the conoid. This shadow 
will be limited by the ray, as A6, thronglh _:_ __ The opposite 
parallel element, T B :s the imaginary shadow, on die interior 
of the conoid, of the portion TT" of the line of stziction. 

77 f 7ii A: z : z : :7r _zz:z: . ::z::.:: z :7t :z:rz:.rj 
ztxtvq of shadow, cast npon the interior of the conoid by the ver- 
tical element at A; it being die horizontal projection of the 



SHADES AND SHADOWS. 67 

intersection of a vertical plane of rays, through this element, 
with the conoid. 

"When the plane of rajs, through the line of striction, does not 
otherwise intersect the conoid, the shadow of this line will fall 
wholly on the horizontal plane. In the case just considered, only 
the portion T 7/ A 7/ casts a shadow on the horizontal plane. 

The remaining shadows on the horizontal plane are cast by the 
curves of shade. They are found by the usual obvious methods 

In this article, only the lower nappe of the conoid has been 
considered. 



68 GENERAL PROBLEMS. 



DIVISION II. 

SHADES AND SHADOWS ON DOUBLE CURVED SURFACES, 

§ 1. — Shades. 

71. Either a line, or a plane, may be placed tangent to a 
double curved surface, at a point of contact, only. Hence a point 
in the curve of shade of such a surface, may be conceived of, 
either as the point of contact of a tangent ray, or of a tangent plane 
of rays. 

72. A tangent plane of rays is, practically, conceived of as con- 
taining some given ray, which, with the point of contact, deter- 
mines that tangent plane. A tangent ray is practically conceived 
of as a tangent to some given, or constructed, section of the given 
double-curved surface. 

73. From the two preceding articles, we have the two follow- 
ing general methods for finding points in the curve of shade of any 
double-curved surface. 

First general method. — That of tangent planes of rays. By any 
of the solutions of the problem of orthographic projections, 
requiring a plane to be drawn through a given line and tangent 
to a double-curved surface, construct a tangent plane through any 
assumed ray. Its point of contact with the given double-curved 
surface, wll be a point in the curve of shade of that surface. 

74:. Second general method. — That of tangent rays. Construct 
any simple section of the given surface, whose plane shall contain a 
ray of light. Draw a parallel ray, tangent to this section, and its 
point of contact will be a point of the required curve of shade. 

Remarks. — a. The first of these methods has no formal illustra- 
tions among the following problems, numerous special methods 
being more convenient in the various cases. 

b. The second general method is convenient, in finding certain 
extreme points, as the highest and lowest, on double-curved sur- 
faces of revolution. It, therefore, has no separate formal illustra* 
tions in the succeeding problems. 



SHADES AND SHADOWS. 69 

c. Each of the following problems is solved by a special method, 
which is convenient for the purpose, but is not necessarily appli- 
cable only to the problem thus solved. 

75. First special method. — That of normal planes of shade. The 
essential principle of this method, which is applicable only to the 
sphere, is, that the plane of the curve of shade — briefly called the 
plane of shade — of the sphere, is perpendicular to the direction 
of the light. 

The peculiarity and convenience of this method is, that it 
requires but one projection of the sphere, as will now be 
shown. 



Pkoblem XXIY. 

To find the curve of shade of the sphere, using only one projection 

of the sphere. 

Principles. — Let the vertical projection, only, be given, and 
let the vertical plane of projection be taken through the centre 
of the sphere. 

Let that plane of rays, which is perpendicular to the vertical 
plane of projection, be called the perpendicular plane of rays, 
and let the one, whose angle with the vertical plane equals that 
made by the light with that plane, be called the oblique plane 
of rays. Let this oblique plane of rays contain the centre of 
the sphere. 

The oblique plane of rays, and the plane of shade, are per- 
pendicular to each other, and to the perpendicular plane of rays, 
hence their traces on the latter plane will be perpendicular to each 
other at the centre of the sphere, and their common vertical 
trace, T, will be perpendicular to the trace of the perpendicular 
plane of rays, and will pass through the centre of the sphere. 

Now let the perpendicular plane be revolved about its trace, 
and into the vertical plane of projection. Then consider the 
great circle, which is the vertical projection of the sphere, as the 
circle of shade, revolved about its diameter, T, into the vertical 
plane. Project any of its points into the trace of the perpen- 
dicular plane. Revolve it into the trace of the plane of shade 
on the perpendicular plane. Counter revolve the latter plane 
to its primitive position, carrying the revolved point. Counter 



70 GENERAL PROBLEM j 

project the point to the parallel plane from which it originally 
came. 

Construction. — PI. IX.. Fig. 26. LPKE is the vertical pro 
jection of the sphere. LK is the vertical projection of a rav, 
and the vertical trace of the perpendicular plane of rays through 
the centre of the sphere. Or, assumed at pleasure, is the 
revolved position of a ray, about LK as an axis, and is also the 
revolved position of the trace of the oblique plane of rays, on 
the perpendicular plane of rays, LK. OS. perpendicular to rO, 
is the revolved position, about the trace LK as an axis, of the 
trace of the plane of shade on the plane LK. POE is the com- 
mon vertical trace of the oblique plane of rajs, 0, and the 
plane of shade, OS, on the vertical plane of projection. It is 
perpendicular to LK, and is the transverse axis of the ellipse 
into which the circle of shade is projected. LBS is the circle 
of shade, after revolution about POE, and into the vertical 
plane of projection. Assume any point 5, project it, at b\ into 
the trace LK. Eevolve it, as at b'b'\ about as a centre, into 
the trace OS of the plane of shade. Counter revolve 6", with 
the plane LK to V" in the primitive position of the plane LK. 
and counter project V" to B, in the plane ab whence it came. 
Then B will be a point in the required projection of the curve 
of shade of the given sphere. 

All the points of this curve may be found as just lescribed, 
but it is not necessary to find more than a quadrant, KE-s, of the 
curve in this way. For. taking parallel plan^ as ab and C^, 
equidistant from LK. the point V" can also be counter projected 
to C, a point of shade opposite to B from the axis s : ;. T_:en r 
and B ; and h and C, etc., will be equidistant from the transverse 
axis PE. 

76. For the practical case in which the projections of a raj 
make angles of 45° with the ground line, the angle, as ?*OL. 
made by the ray with the vertical plane of projection, will be 
35 : -16 ; (21). OS, being perpendicular to Or. KOS will be an 
angle of 54 C ^L4'. This angle may easily be constructed, then it 
will not be necessary to draw Or. Thus, take on OK any dis 
tance, 0-s, and find the hypothenuse of a right angled triangle, 
each of whose other sides equals Os. On a perpendicular, a« 
sS, at s, lay off this hypothenuse, then SO, the new hypothenuse 
5 found, will make the required angle of 54 = — 44' with OK, 
and will therefore be the true position, after revolution of the 



SHADES AND SHADOWS. 71 

plane LK, of the trace of the plane of shade upon the plane 
LK. 

77. Second special method. That of the intersection of the plant 
of shade with the given double curved surface. This method can 
be applied to all the double curved surfaces of the second 
order (Des. Geom.) for the curve of contact of a cylinder, or 
cone, of rays, with such a surface, is a plane carve of shade. This 
method consists principally in making an auxiliary rectilinear projec- 
tion of the curve of shade, which may be done by taking an auxiliary 
vertical plane of projection perpendicular to that plane of shade. 



Problem XXV. 

To find the curve of shade on an ellipsoid of revolution. 

Principles. — Let the ellipsoid be a prolate spheroid (Des. 
Geom. 219) and let its principal axis be vertical. Any auxiliary 
horizontal plane will then cut a horizontal line from the plane 
of shade and a circle from the ellipsoid. 

The two points of intersection of these two lines will be 
two points in the curve of shade. 

Construction. — PI. IX., Fig. 27. Only half of the ellipsoid is 
here shown. A — A'D' is its semi-principal axis. BO — B'C is 
the diameter of its greatest horizontal section. E"F", parallel 
to LA, the horizontal projection of a ray of light, is the ground 
line of a new vertical plane of projection, parallel to the light. 
The projection of the semi-ellipsoid on this plane is a semi-ellipse 
equal to B'D'C 7 . Having noted L and L', two projections of 
any point in the ray LA — L'A', project A at A", and L at L", 
making JJ'h" equal to Uh ' . Then L"A" is the projection of 
the light on the new plane of projection. Next, draw the tan- 
gent ray at T 7/ , and T^A" will be the auxiliary rectilinear 
projection of the curve of shade. Now the auxiliary horizontal 
plane, m"n", taken at pleasure, cuts from the plane of shade the 
line a" — ah, and from the ellipsoid, the circle m"n" — anb. 
These intersect at a,a" and 6, a", two points of the curve of 
shade. On the primitive vertical projection, these points are 
projected at a' and b\ on a line a'b\ as far from B'C as a" is 
from the ground line E"F." 



72 GENERAL PROBLEMS. 

Similarly other points are found. A" is projected at S and V, 
and then at S' and V. T" is likewise projected at T and T'! 
Drawing a tangent, parallel to I/A 7 , we find t' ,i, on the meridian 
curve parallel to the vertical plane whose ground line is B'C. 

Only the portion t'b'Y'C of the shade is visible in vertical 
projection. The boundary of the invisible portion of the shade 
is dotted. 

Bzmark. — The point T" is exactly constructed on the princi- 
ple that the diameter of an ellipse, through the point of tangency, 
bisects all the chords of the ellipse parallel to the tangent. 
Hence take any two chords, parallel to I/'A", bisect them, and 
draw a line through their middle points. Such a line will meet 
the circumference at T", the point of tangency of a tangent 
parallel to L"A". 

78. Third special method. This is essentially the second general 
method (74) but modified so as to be most conveniently applica- 
ble to the construction of the curve of shade on an ellipsoid of 
three unequal axes. 

A plane of rays passed through the mean axis, will be per- 
pendicular to the plane of the longest and shortest axes. Make 
the ellipse, E, cut from the ellipsoid by this plane, the base of a cylin- 
der whose right section shall be a circle. Such a cylinder may be 
considered as the projecting cylinder of the ellipse ipon the plane of 
right section of the cylinder, taken as a new plane of projection. 
Other parallel planes of rays will cut ellipses, similar to E, from the 
ellipsoid. Their projections on the new plane of projection will 
therefore be circles, to which tangent rays may be drawn, which will 
give projections of points of the curve of shade. 



Problem XXVI. 

To find the projections of the curve of shade on an ellipsoid of three, 

unequal axes. 

Principles. — Let the longest axis be vertical, and the plane of 
the longest and shortest axes, parallel to the vertical plane of 
projection. To locate the projecting cylinder described in (78) 
make the centre of the ellipsoid the centre of an auxiliary 
sphere, whose radius shall be equal to the semi-mean axis of the 



PL IX 




\ 



SHADES AND SHADOWS. 73 

ellipsoid ; then the cylinder will be tangent to this sphere in a 
circle, which may be considered as a circular projection of the 
elliptical base of the cylinder. 

Any plane, parallel to the circle just described, may be taken 
as a new horizontal plane of projection, on which a new hori- 
zontal projection of a ray of light must be constructed. 

Construction. — PI. IX., Fig. 28. 0,0' is the centre of the 
ellipsoid. Let / D / be the length of the semi-mean axis, whose 
vertical projection is 0'. A circle, with radius O'D' and centre 
CK, is the vertical projection, partially shown, of the auxiliary 
sphere above described. 

LO and L / / are the projections of a ray of light. Then 
L'F' is the vertical trace of a plane of rays, perpendicular, here, 
to the vertical plane of projection, in being perpendicular to the 
plane of the longest and shortest axes. J'F' is the vertical pro- 
jection of the ellipse contained in this plane of rays, and tan- 
gents, from J' and F', to the vertical projection of the auxiliary 
sphere, as at e', are the extreme elements of its projecting cylinder. 
Take, now, any new ground line, MN, parallel to OV, and make 
CO" perpendicular to it, and make 0" and equidistant from 
their respective ground lines, MN and MP. Likewise make L" 
in a perpendicular from L' to MIST, and as far from MN as L is 
from MP. Then L"0" is the new horizontal projection of a ray 
of light. Now the circle with centre 0" and radius 0"F (=0V) 
is the horizontal projection of the section J'F', and its diameter 
frij perpendicular to L'W, determines precisely the points of 
contact, f and n, of rays tangent to the section J'F 7 — JnFf. 
These points being projected at/ 7 and n\ give fj' and n^n' for 
two points of the required curve of shade. 

By drawing the tangent rays at t' and T', we find the line t ; T\ 
which bisects all the chords, as K'B', parallel to the tangents at 
t' and T v . These chords are the vertical projections of ellipses, 
whose centres, as m', are in the line t'T', and whose horizontal 
projections are circles having their centres in Tt, the horizontal 
projection of t ,T Y\ 

Thus, m is the centre, and mK the radius, of the horizontal 
projection of K'B', and Job, its diameter perpendicular to I/'O", 
determines the points of shade, h and b, vertically projected at 
h' and V. In the same way g,g' and h,~ti are found. 

Through the points now found, the projections of the required 
curve can be sketched ; showing that it will be included between 



74 GENERAL PROBLEMS. 

tangents (not shown) as at u and y. tangent also to the vertical 
projection, t'f'T'u. and parallel to O'O". 

The horizontal projection of the curve of shade is also tangent 
at u and y. to the ellipse, not shown, whose minor axis is Oa — 0', 
and whose longer axis is the distance, on KF, between two 
lines, perpendicular to it, and tangent to the ellipse T'C'A'J'; 
and the former ellipse is the new horizontal projection of the 
ellipsoid. 

The visible portion of the shade is shaded in vertical pro- 
jection. 

79. Fourth special method. This is essentially the special 
method of (51;, but modified so as to be also applicable to the 
ellipsoid of three unequal axes. It is a property of these ellip- 
soids — employed in the preceding article — that a set of parallel 
sections of any one of them, is composed of similar ellipses. 
Add to this, as just previously shown, that a diameter can 
obviously be found, in the plane of the longest and shortest 
axes, which shall be equal to the mean axis. The plane of this 
diameter and mean axis will evidently intersect the ellipsoid in a 
circle, and so will all parallel planes. Xow by taking a plane 
of projection — treated most conveniently as a new horizontal 
plane — and made parallel to these circles, we can readily con- 
struct the elements of shade on a series of auxiliary tangent 
cones having these circles for their bases. The intersection of 
the elements of shade with the circles will be points in the curve 
of shade of the ellipsoid. 

The construction is left for the student to make. 

80. Fifth special method.. This is the method of projections of 
rays on meridian planes. Its essential principle is, that the point 

of contact of a tangent line to anv meridian curve of a surface 
of revolution, and parallel to the projection of a ray on the 
plane of that curve, is a point of the curve of shade of that sur- 
face. For, a point of shade is the point of contact of a tangent 
plane of rays (73). But such a plane is perpendicular to the 
meridian plane through its point of contact. Hence its trace on 
the meridian plane is the projection of a ray upon that plane, and 
is also a tangent to the meridian carve, at the point of contact 
of the tangent plane ; that is, at a point of shade ; which prove* 
the principle above stated. 



SHADES AND SHADOWS. 75 

The general operations of solution under this method are 
exemplified as follows : 



Pkoblem XXVII. 
To construct the curve of shade upon a torus. 

Principles. — Conceive of a rectangle, with one of its shortex 
sides made the diameter of a semi-circle. If the entire plant 
figure, thus formed, be revolved about the opposite side of the 
rectangle, the solid generated will be a torus. 

If the axis of the torus be supposed to be vertical, the points 
of shade on its greatest horizontal section will be the points of 
contact of two parallel and vertical tangent planes of rays. The 
points of shade on the meridian curve parallel to the vertical 
plane, will be the points of contact of two planes of rays per- 
pendicular to the vertical plane of projection. 

Hence, in finding the four points just described, we make a 
rudimentary application of the first general "method (73). Other 
points of shade are mostly found by the fifth special method (SO). 

Constructions. — 1°. To find the four points on the visible bounda- 
ries of the torus. PI. IX., Fig. 29. Let ATB* and A'L'B' be 
the projections of the torus, and RL — B/I/ a ray of light. Then, 
by drawing T£, perpendicular to RL, T and t! will be determined 
as the horizontal projections of the points of contact of vertical 
planes of rays, tangent to the greatest circle of the torus at T,T' 
and t/. 

Likewise, at C and D r , the centres of the semi-circular ends 
of the elevation, draw Q'n' and DV, perpendicular to RT/, and 
project n' at w, and c' at c, then n,n' and c,c' will be the points 
of shade on that meridian section of the torus, which is parallel 
to the vertical plane of projection. 

2°. The highest and lowest points. These are in the meridian 
plane of rays KL. Revolve this plane, with the ray RL — RT/, 
about a vertical axis at L, till parallel to the vertical plane of 
projection. Then r"L — R"L' is the revolved position of the 
ray RL— R'L'. Through C and D', draw CV and DW per- 
pendicular to R"L", and project u' at u and a' at a. Then a,a f 
and u,v! are the revolved positions of the highest and lowest 
points. In the counter revolution, a, a' returns to a'V", and u,u' 



76 GENERAL PROBLEMS. 

to u"u ff 'i which are the highest and lowest points of the required 
curve of rfhade. These points are thus found by a simple appli- 
cation of the second general method (74). 

3°. To find intermediate points. Assume any meridian plane, 
as d"\js!\ project the ray KL — R'L' upon it by projecting lines, 
as Rr — RV, perpendicular to cf 'Le". L,I/, being in the assumed 
plane, is its own projection on that plane, and rL and r'U are 
the projections of the projection of the ray, RL — R'L', upon the 
plane d"Le". 

If, now, the plane d"W be revolved about a vertical axis at 
L, till it is parallel to the vertical plane of projection, the meri- 
dian section contained in it will coincide with the vertical pro- 
jection of the torus, and the projected ray rL — r'\J will appear 
at R""L— R'"L'. From C and D', draw lines, Q'd' and DV, 
perpendicular to the revolved ray R ///7 L — R^'L', and project d' 
and e' at d and e. Then, in the counter revolution of the meri- 
dian plane, d,d' will return in a horizontal arc to d" ,d'" . and e,e' 
to e'\e"\ which are points of the required curve of shade. 

Other points can be similarly found. 

81. The special method of (51) may be applied to any double 
curved surface of revolution, but for the sake of completeness 
of illustration, and to show how it would be applied in finding 
the carve of shade on a warped hyperboloid, it is here applied 
to a concave double curved surface of revolution, analogous in 
form to that warped surface. This method will therefore be next 
described in immediate connexion with the " construction " of the 
following example : 



Problem XXVIII. 
To find the curve of shade on a piedouche. 

Principles. — A Piedouche, PI. X., Figs. 30, 31, is a little orna- 
mental pedestal, used for the support of a bust, a statuette, etc. 

Conceive of an ellipse, whose axes are in a vertical plane, but 
oblique to the horizontal plane. Suppose a vertical line, L, in 
the plane of the ellipse, and so placed as to converge downwards 
towards its minor axis. Then let that arc of the ellipse which is 
included between horizontal tangents, and convex towards L, be 
revolved about L as an axis, and a concave double curved sur- 



SHADES AND SHADOWS. 77 

face of revolution will be formed, which, will be the principal 
surface of the piedouche. The lower base will evidently be 
larger than the upper, and both may be made the bases of short 
cylinders having L for their axis. See PL X., Fig. 30, which 
shows how the generating semi-ellipse, AEB, may be formed 
from a semi-circle, as ACB, where DE, etc.=DC, etc. 

The elliptical generatrix may be replaced by a compound 
curve, composed of circular arcs, tangent to each other, and with 
different radii. 

Remark. — The piedouche formed by the second method will 
answer every graphical purpose, but is less ornamental, since a 
circle is a monotonous curve, its points being equidistant from 
the centre ; while the gradually varying distance of the points 
of an ellipse from its centre gratifies the natural love of variety. 
While dwelling on a point of taste, it may be added that the 
generatrix should begin and end on horizontal tangents, because 
it thus suggests completeness ; and this, because horizontal boun- 
daries imply stability ; and this, again, because gravity, in refer- 
ence to whose action the piedouche is to be stable, acts vertically. 

Constructions. — 1°. Of the piedouche, and of some preliminary 
points of shade. PI. X., Fig. 31. Assume the bases, H / C / and 
F'Gr", of the double curved portion of the piedouche. At F r , 
take any vertical height F'E', greater than half the distance 
between the assumed bases. Through E', draw the line ET', 
parallel to the bases, and, from C, let fall CT)', perpendicular to 
E'F. Then, with E'F' as a radius, describe the quadrant from 
F' to E ; F, and with D'O' as a radius, describe the quadrant from 
ET to C. 

The compound curve, thus formed, will be the meridian curve, 
whose revolution about a vertical axis, A — T'A', will generate 
the double curved portion of the piedouche. Its cylindrical 
bases are, as shown in vertical projection, above and below the 
double curved portion. These bases are horizontally projected 
in the concentric circles, A — HYB, and A — GWF. 

In applying the method of (51) some of the assumed circles 
of contact of the auxiliary surface may prove to be beyond the 
limits of the curve of shade, so that it is best to construct the 
highest and lowest points of that curve first, as in (Prob. 
XXYIL). 

Let A J — A' J' be a projection of a ray of light. Eevolve 
this ray about the axis A — T'A' to the position AJ" — A 'J'" 



73 GENERAL PROBLEMS. 

Draw IV and EV" perpendicular to A.' J'", then u' and /", 
horizontally projected at u and r", will be revolved positions of 
the highest and lowest points. In the counter-revolution, these 
points return to UjU' and r,r', which are their real positions. 

By drawing IV and E V, perpendicular to AM', we find a', a 
and n',n, the points of tangency of planes of rays perpendicular 
to the vertical plane of projection. That is, aa! and nn' are the 
points of shade on the meridian curves forming the vertical pro- 
jection of the piedouche, and contained in the meridian plane 
HAF, parallel to the vertical plane of projection. See the first 
general method (73). 

2°. Of intermediate points of shade. First illustration of (81) 
Principles. Assume any horizontal section of the piedouche — 
such a section being circular because the axis is vertical — and 
make it the circle of contact, C, of a tangent cone, whose ele- 
ments will therefore be tangent to the meridian curves of the 
piedouche. Drawing planes of rays, tangent to this cone, we 
find its elements of shade. The two points of intersection of 
these elements with the circle of contact, C, will be points of 
shade on the piedouche (51). At the circle of the gorge, the 
cone will become a tangent cylinder. 

Construction. — Assume any horizontal plane, KT/, PI. X, Fig. 
31, between the highest and lowest points. At either point, as 
I/, of its intersection with the meridian boundary of the vertical 
projection of the piedouche, draw the tangent L'S'. This tangent 
will be an element of the auxiliary tangent cone, whose base is 
Z'l/. 

The circle, with radius Ah, equal to XT/, is the horizontal 
projection of the circle of contact, or base, of the cone, and A,S' 
is the vertex of this cone. The ray AK — S'K', through the 
vertex A,S', pierces the plane, KT/, of the base at K',K. Hence 
K& is the horizontal projection of the trace on the plane KT/, 
of a plane of rays, tangent to the auxiliary cone. Project k at h' ; 
then, as Tcjc' is the intersection of the element of shade, Ah — 
S'h\ of the cone, with the circle of contact, Z'L 7 , it is a point, of 
tangency of the plane of rays, K&, with the piedouche ; that is, 
\~k' is a point in the curve of shade of the piedouche. But as 
two planes of rays can be drawn, tangent to the cone, each cone 
will give two points of shade. Thus, by drawing the chord kt, 
perpendicular to KA, we find the point of shade. t,t', both of 
whose projections are invisible. 



SHADES AND SHADOWS. 79 

Likewise, by assuming a circle of contact h'c\ whose radius, in 
horizontal projection, is AQ, equal to half of b'c f , and which is 
the base of a tangent cone whose vertex is W, we find Q,Q / and 
R,B/ for two more points of the curve of shade. 

Finally, at the circle of the gorge, f'm\ horizontally projected 
with a radius Ah, equal to half of f'm\ the auxiliary cone becomes 
a cylinder. Hence, by drawing the diameter gh, perpendicular 
to AK, the direction of the light, we find the two points of shade, 
g,g' and h,h\ 

Remark. — The elements of shade, JcA and £A, produced in 
the direction JcA and tA, will meet the base of a cone, tangent 
to the lower nappe of the piedouche, and having the same inclina- 
tion that the element L / S / has, and will therefore give two more 
points of shade, which can be easily found. 

3°. Second illustration of the method of (51). 

Principles. — Make any circle, between the highest and lowest 
points of shade, the circle of contact of an auxiliary sphere, tan- 
gent to the piedouche internally. The meridian plane, parallel 
to the vertical plane of projection, and the plane of the assumed 
circle of contact, which is horizontal, may be taken in this con- 
struction as the planes of projection. But the plane of the curve 
of shade of the sphere is perpendicular to the direction of the 
light, hence its traces will be perpendicular to the projections of 
a ray. The vertical trace will pass through the centre of the 
sphere, and its intersection with that diameter of the circle of 
contact which lies in the meridian plane just mentioned, is a point 
of the horizontal trace of the plane of shade of the sphere. The 
intersections of this horizontal trace with the circumference of 
the assumed circle of contact, will be two points of the curve of 
shade of the piedouche, being the points in which the curve 
of shade of the sphere intersects the circle of contact of the sphere 
and piedouche (51). 

Construction. — PI. X, Fig. 31. Let WW — d'Ne be the assumed 
circle of contact. Produce WW to T' in the axis of the piedouche ; 
then T'N' is the radius of the auxiliary sphere, whose circle of 
contact with the piedouche is M'N'. Then T'O', perpendicular 
to A'J', is the vertical trace of the plane of shade of the sphere, 
and (y,0 is a point of its horizontal trace, eOd. The points e,e' 
and d,d\ as explained above, are therefore points of the curve of 
shade of the piedouche. Through the points now found, the 
curve can be sketched. It is wholly invisible in horizontal pre- 



30 GENERAL PROBLEMS. 

jection. That part is visible in vertical projection which is in 
front of the meridian plane GrAF. 

Remarks. — a. The curve of shade on a sphere may be found as 
in Prob. XXIV., in Prob. XXV., in Prob. XXVII. Its points 
may also all be found as are the highest and lowest on the 
piedouche or torus, since a set of vertical planes of rays would 
cut the sphere in circles ; or they may be found by auxiliary 
tangent cones, as in (2°.) 

b. Since the piedouche is a surface of revolution, its curve of 
shade may be found as in the problem of the torus. Likewise the 
curve of shade of the torus may be found by the methods just 
employed. In fact, these methods are especially applicable to the 
interior half of the annular torus, a surface generated by the 
revolution of a circle about a line exterior to it, but in its plane. 

c. By inspection of the figure of the piedouche, the following 
curious property may be observed. Near the points K,R/, Q,Q', 
1,1', and 2,2', tangent rays may be drawn to the projections of 
the curve of shade at the two projections of the same point. 
Hence such tangent rays are really tangent to the curve of shade 
in space. These tangents evidently include the greatest hori- 
zontal chords of the curve of shade, one on each nappe of the 
piedouche. 

d. It may be shown in several ways, that there must be such 
points of maximum width of the curve, at which tangent rays 
may be drawn to the curve. This may be done, in an elementary 
manner, as follows. Suppose all points of the curve of shade to 
be found by the general method of (74) employed in finding the 
highest and lowest points of shade. A vertical plane of rays 
near the gorge, will cut a curve similar to A!'W, PI. X., Fig. 32, 
to which two tangent rays can be drawn, giving two points of 
shade, T and T'. Consideration of the form of the piedouche 
will show, that, as the vertical secant plane recedes from the 
gorge, the curve, A^B", will become flatter, the points T and T / 
will approach each other, and will finally unite, as at T, on A'B' ; 
T being a point of inflexion, or change of curvature. 

If the cutting plane recedes still further from the gorge, the 
curve will be, as at AB, so flat that no tangent ray can be drawn 
to it ; hence the point T marks the position of the greatest width 
of the curve. 

e. Moreover, the ray is tangent, at T, to the curve of shade 
itself, for, on that position of A"B" which is consecutive with 



rl. A 




PI X 




SHADES AND SHADOWS. 81 

A'B', T and T' will be consecutive points of shade, and their 
tangents consecutive parallels. On A'B', therefore, T and T" 
unite, and their separate tangents coalesce, forming a tangent to 
the curve of shade at T. 

f Those of the secant planes which intersect the gorge, cut 
the piedouche in curves analogous to the meridian curve. 

g. Furthermore, the points, as T, Fig. 32, near 1, 2, E, etc., 
on PI. X., Fig. 31, are the limits between that part of the curve 
of shade which really exists on the exterior of a solid opaque 
piedouche, and on a transparent one, open at the top, and con- 
sisting merely of a surface without thickness. All points, as T, 
on the upper nappe, and similar ones at such points at T', on the 
curves like A^B" cut from the lower nappe, are imaginary points. 
on the interior side of the hollow surface. Only such points as 
T 7 , on curves shaped like A^B" and cut from the upper nappe, 
and such as T, situated on curves cut from the lower nappe, are 
points in which rays in space are truly tangent to the external 
surface of the piedouche. The construction of a few intersec- 
tions of the piedouche by vertical planes of rays will make these 
statements clear. 

h. There is a direct construction, given by M. Leroy, of the 
points, as T, of maximum width of the curve of shade, but it is 
extremely tedious and comparatively unimportant, being, in the 
graphical point of view, only approximate. Hence it is not 
here given. Moreover, it is unimportant, for the further reason 
that these points are found with sufficient accuracy in sketching 
the curve by hand through a considerable number of other con- 
structed points. 



§ n. — Shadows. 

82. The direct method of finding shadows on double curved 
surfaces, is essentially the same as for finding shadows on other 
surfaces. Pass a plane of rays through any point, P, of the line 
casting the shadow, and construct the intersection, I, of this plane 
with the given double curved surface. The point, where a ray through 
P meets the line, I, is the shadow of P on the double curved surface. 

It being desirable, for graphical convenience, that the line .1 
should be a circle, it is evident that the foregoing direct method 
is always conveniently applicable only to spheres ; and to other 

6 



82 GENERAL PROBLEMS. 

surfaces of revolution, only when so situated in reference to the 
light, that planes of rays may be made to cut them in circles. 
As these conditions do not generally occur, the shadows on 
double curved surfaces are mostly found by a variety of indirect 
and special methods, which will be sufficiently understood from 
the following problems. 



Problem XXIX. 

To construct the shadow of the front circle of a niche upon its spheri- 
cal part. 

Principles. — This problem, enunciated more as an abstract one, 
is this : To construct the shadow of a vertical circle upon the 
interior of a hemispherical surface of which the given circle is 
the front edge. 

83. If a straight line be moved, so as to remain constantly 
parallel to its first position, and to rest on the circumferences of 
two great circles of a sphere, it will generate a cylinder, of which 
those circles will be equal oblique sections. All the elements of 
this cylinder will intersect the sphere, except the two which are 
tangent to it at the extremities of the common diameter of the 
two great circles. Hence we have the principle, that a cylinder 
which intersects a sphere in a great circle, intersects it also in a second 
great circle. The right section of this cjdinder will be an ellipse, 
equally inclined to these great circles, and whose transverse axis 
is their common diameter. 

It now follows, that the cylinder of rays passing through the 
front edge of the quarter sphere, will intersect its interior in an 
arc of a great circle of that sphere, which arc will be the shadow 
required. 

1°. The first solution, based on the foregoing general princi- 
ples, is made by the direct general method (82), and depends imme- 
diately on the following — 

Principles. — A plane of rays, perpendicular to the vertical plane 
of projection, will cut a circular arc from the spherical part of the 
niche, and a ray from the cylinder of rays, whose intersection will 
be a point of the required shadow. 

Construction. — PI. X., Fig. 33. In this construction, three 
planes of projection are u^ed. 



SHADES AND SHADOWS. 83 

The plane of the front of the niche is taken as the vertical plane 
of projection. The plane of the upper base of the niche is taken 
as the horizontal plane of projection, AC is therefore the ground 
line for these planes, and the spherical quadrant of the niche is 
in their second angle. D'H', parallel to the vertical projection, 
DO, of a ray of light, is the trace, or ground line, of the auxili- 
ary plane of projection. D'H' is in the vertical plane, and the 
auxiliary plane is revolved about it, into the plane of the paper, 
so that the part in front of the vertical plane falls to the right 
of D'H'. 

DO — J)"p is a ray of light. When a point is projected on 
two planes, each perpendicular to a third, its projections on those 
planes are at equal distances from their respective ground lines, 
hence, making O'p' equal to Op, p' is the auxiliary projection 
of the point 0,p of the ray DO — D"j9. D is projected at D', 
therefore D'jp' is the projection of the ray, DO — D"j9, on the 
auxiliary plane. 

The plane of rays, whose vertical trace is DO, cuts from the 
spherical part of the niche, produced, the semicircle DOc — 
D'T'H' and from the cylinder of rays (83), the ray, DO — T)'p\ 
which pierces the semicircle at c', which, being projected back 
at c, gives c,c 7 as one point of the required shadow — produced. 

Likewise, any other parallel plane of rays, as ah, cuts from 
the niche t4ie semicircle ah — a'h'f", and from the cylinder of 
rays, the ray ah — a'h', giving hft for another point of shadow. 

As N'T' is the ray which is the tangent element (83) of the 
cylinder of rays, T' is where the shadow begins, and T'Ac is the 
quadrant of shadow, on the spherical part of the niche, pro- 
duced. 

2°. The second Solution is made by the special method of (42). 

Construction. — PL X., Fig. 34, OHB— O'A'B' is the upper 
base of the semi-cylindrical part of the niche. CAB — C'A"B' 
is the front edge of its spherical part. 

By the first solution, T,T', where a plane of rays, N'T', perpen- 
dicular to the vertical plane of projection, is tangent to the 
spherical part, on its front edge, is the point where the required 
shadow on the spherical interior begins. 

To find another point, take any auxiliary plane, parallel to the 
front face of the niche. EH may be assumed as the trace of 
such a plane, on the plane C'A'B'. The ray AE — A'E', 
through the centre of the front semicircle, pierces the vertical 



8-4 GENERAL PROBLEMS. 

plane EH, at EjE', which, is therefore the centre of the shadow 
of that semicircle on the plane EH. With E' as a centre, and 
a radius equal to A'C (33), describe the arc which intersects the 
semicircle Wh' t cut from the quarter sphere by the plane EH, 
at h! ; whose horizontal projection is A, on the trace EH. h,h' is 
therefore a required point of shadow. 

The point y,y' is similarly found, on a similar vertical secant 
plane, aY. 

3°. Construction of the point where the shadow leaves the spherical 
part of the niche. 

This point, b,b\ may here be found, without reference to the 
cylindrical part of the niche. 

Remembering that the plane of shadow contains a great circle 
of the spherical part, the diameter A'T' is its trace on the front 
face of the niche, and therefore A' A is one point of its trace on 
the base, CHB, of the spherical part. That trace intersects the 
semicircle CHB at the point required. Here then is given one 
trace, A'T', of a plane, and a point, as y,y', in that plane, to find 
its other trace. ]Sow ya — y'a' is a line through y,y\ and paral- 
lel to the trace AT, and therefore is a line of the plane of 
shadow. This line pierces the plane C'A'B' at a',a, giving Aab 
for the required trace on C'A'B', and b,b' for the required point 
at which the shadow leaves the spherical part of the niche ; it 
being the intersection of the plane of shadow with the base 
line CHB. 

4°. Third Solution. — This is made by the special method of 
(43) which, in its present application, depends immediately on 
the following — 

Principles. — Knowing, in advance, from (83) that the curve 
of shadow is a plane curve, we can find its rectilinear projection 
on a plane of projection, taken perpendicular to the plane of 
shadow. Then, having this rectilinear projection, both projec- 
tions of particular points in it can be found as in (43). 

Construction. — PL X., Fig. 33. D'H' is the trace, on the plane 
of the face of the niche, of a plane perpendicular to the plane 
of shadow. All the planes of projection are, moreover, the 
same here, as in the first solution. 

Having found, as before, c', we have OT'Y for the vertical, 
and OV for the auxiliary, trace of the plane of the shadow, the 
latter trace being also the projection of the shadow on the auxili- 
ary plane. Any semicircle, as a'h'f ' ', lying in the spherica] 



SHADES AND SHADOWS. 85 

surface, or any ray, as a/h\ will cut from the given projection of 
the shadow a point, as A', whose other projection, A, is found by 
drawing h'h, perpendicular to the ground line D'H', and noting 
its intersection, h, with the vertical projection, ah, of the same 
semicircle, a'lif", or ray o/h'. 

Discussion. 

Fiest. — If it were not known already that the shadow on the 
spherical part of the niche is a plane curve, it could be proved by 
reference to PI. X., Fig. 33. For, O'D' and V are equal, hence 
OVD' is an isosceles triangle. But a'h' being a chord, parallel 
to DV, in a semicircle a'h'f", concentric with DVH', it follows 
that O'a'h' is an isosceles triangle also, and is similar to OT)V. 
Then, as OV and O'D' coincide in direction, 0'h ; and OV like- 
wise coincide, and OTiV, the projection of the shadow on the 
auxiliary plane, is a straight line, which shows the shadow itself 
to be a plane curve. 

Second. — Another construction of the point, e, where the sha- 
dow leaves the spherical part of the niche, may here be given. 
C'B" is the auxiliary projection of a line, perpendicular to the ver- 
tical plane of projection at 0. Its horizontal projection would be 
a perpendicular to the ground line, AC, at C ; but, to avoid con- 
fusion of the figure, a similar line is shown at AB. Producing 
OV to B ;/ , gives B" as the point in which the plane of shadow 
intersects C'B". Now make AB equal to C'B", and OB will be 
the horizontal trace of the plane of shadow, after revolving 180° 
about OU, in the horizontal plane, as an axis. The point m is 
then the revolved position of the intersection of the horizontal 
trace of the plane of shadow, with the horizontal semicircle, 
AUG, of the niche. This intersection is the point sought. Then 
the horizontal projection, mn, of the vertical arc of counter- 
revolution, gives n for the primitive horizontal projection of this 
point, and ne, perpendicular to the ground line, AC, gives e, the 
vertical projection of the same required point of shadow. 

Third. — The point e can again be found, but considered still 
as the intersection of the horizontal trace of the plane of shadow 
with the horizontal semicircle, AUO, of the niche. This is done 
by viewing this trace as the intersection of the plane of shadow 
with the plane of the horizontal circle AUC. Then, after show- 
ing the traces of both of these planes on the planes of projection 



86 GENEKAL PEOBLEMS. 

whose ground line is D'H', their line of intersection can b€ 
found. 

The traces of the plane of shadow are OT'Y, on the vertical 
plane, and O'B", on the auxiliary plane D'H'. The traces of the 
plane of the base of the niche are T'O'Y, on the auxiliary plane, 
and OV, on the vertical plane ; for, thus considered, D'H' is 
supposed to be the trace of the plane DO itself, after being trans- 
lated, parallel to itself, to the position D'H', and then revolved, 
as before described. O'e" is thus parallel to OC, being only a 
transferred position of OC. 

The intersection of the plane of shadow and the plane of the 
base is now projected in OV and O'e". The arc/"/" represents 
a revolution of the plane of the base, together with the intersec- 
tion just noted, about 0'— T'O'Y as an axis, and into the trans- 
lated position, DTF, of the plane of rays DO. The base then 
appears in DVH', and its intersection, O'e" — OV, with the 
plane of shadow, at O'f. At e',e'" is therefore the revolved 
position of the point of shadow on the base of the niche. In 
counter-revolution, e'e'" returns to e"E, from which is projected, 
at e, the required projection of the required point of shadow. Or, 
e f may be first counter-projected at e"", and then counter-revolved, 
as shown by the arc e""e % which gives, again, the same required 
point, e. 

Fourth. — Considering D'EP, again, as the ground line of the 
original position of the auxiliary plane of projection, e may again 
be found by the intersection of the projection, OV, of the shadow 
on the spherical quadrant, with the auxiliary projection, C'ET 7 , 
of the base, AUC, of the niche. C'ET' is a quarter ellipse on 
the semi-axes O'T' and O'C, C being the projection of C. 

Fifth. — A little consideration will show that CO' : C'H':: 
Er : Efi'. 

Hence O'T', C'E, and We' all meet at I. Hence, from any 
point, I, on O'O, draw lines to C and H'. From the intersec- 
tion, as e', of IH' with the semicircle D VH', draw a perpendicu- 
lar to O'O ; its intersection, as E, with IC will be a point of the 
ellipse. 

Thus, after finding e', as above, we may find E, and hence <?, 
by drawing HVI, and then IC which will intersect OV at E ; 
which is projected by E<?, perpendicular to D'H' at e. 

Moreover, we have here a construction of the ellipse by homo- 
logous secants, analogous to the homologous tangents which may 



SHADES AND SHADOWS. 87 

be drawn at e' and E, and which will meet CO at a common 
point. 

Sixth. — Finally, e may be approximately fcand as the point 
where the entire quadrant of shadow, TAc, on the spherical pan 
produced, crosses AC. (See 42.) 

Seventh. — To construct the tangent line at the point, e. 

Principles. — At the point e, the required line is tangent both to 
the spherical branch, ehT\ and to the cylindrical branch of the 
shadow. For, the tangent, L, to the spherical branch, is the 
intersection of a tangent plane, P, to the sphere of the niche at e, 
with the tangent plane, P', to the cylinder of rays, along the ele- 
ment eE'. The tangent, I/, to the cylindrical branch, is the 
intersection of the same tangent plane, P', to the cylinder of rays, 
along the element eE r , with the tangent plane, P", along the 
element, eB", of the cylinder of the niche. Now, since this 
cylinder and the sphere are tangent to each other on AUO, the 
planes, P and P", coincide, hence the tangent lines, L and I/, 
coincide also, forming a common tangent to the two curves of 
shadow which coalesce at e. 

Again : The curve ehT' is a plane curve, and the tangent to 
a plane curve of intersection is the intersection of the plane of 
that curve, with a tangent plane to the surface on which it lies. 
Hence, finally, the required tangent line is best found as the 
intersection of the tangent plane to the cylinder of rays, along the 
element eE, with the plane of the curve of shadow. 

Construction. — PL X., Fig. 33. E'Y, tangent to the front of 
the niche, or base of the cylinder of rays, at E', is the vertical 
trace of the tangent plane to the cylinder of rays, along the ele- 
ment eW. OT'Y is the vertical trace of the plane of shadow. 
Then Ye, joining the intersection of these traces with the given 
point e, which is common to both planes, is their intersection, or, 
the required tangent line. 

Eighth. — Having now, by previous constructions, the tangents 
at T', h\ and c, PL X., Fig. 34, where c represents the upper end 
of the vertical shadow of D'C, it is easy to construct, according 
to elementary plane geometry, the curved shadow by approxi- 
mate circular arcs. Thus, from T' to V may be made part of an 
oval of three centres, having A'T' and A / E / for its axes ; and h'c 
may be a single arc, tangent to the tangents at V and c, or, if 
necessary, it may be an arc of a compound curve having the same 
tangents, and a horizontal line at c for one axis. When the 



88 GENERAL PEOBLEMS. 

projections of the light make an angle of 45° with the ground 
line, a single arc can be drawn through h' and tangent at T', and 
another, through b' and tangent at c ; which will nearly coincide 
with the shadow. 



Problem XXX. 

To find the shadow of the upper circle of a piedouche upon its 

concave surface. 

Principles. — The highest and lowest points of this shadow are 
found by the direct general method (41). Other points are found 
by the special method of (42). 

Construction. — PL XI., Fig. 35, shows the surface represented 
as in PL X., Fig. 31. Its projections need not again be described. 

1°. The highest and lowest points of shadow. 

AK is the horizontal trace of a meridian plane of rays. Since 
the piedouche is a surface of revolution, if we revolve AK about 
the vertical axis at A, till it becomes parallel to the vertical plane 
of projection, the meridian curves contained in it will coincide 
with Wd'" and B'D'C, which are seen in the vertical projection 
of the piedouche. 

F,F' is the revolved position of the point whose horizontal 
projection is F" and which is cut by the plane of rays from the 
upper circle casting the shadow. Then the revolved position of 
a ray through F,F', will intersect F'cT'E' and B'D'C, respec- 
tively, in the revolved positions of the highest and lowest points 
of shadow, viz. at d'",d", and r'"r", in the meridian plane FAC. 

The revolved ray, Y'd'"r'", is parallel to AV", which is found 
by revolving the ray Ac — AV to the position Ac" — AV", 
parallel to the vertical plane of projection. 

In the counter-revolution, d",d' n and r'\r'" return, respectively, 
in the horizontal arcs, d"d — d'"d\ and r'r — r"V, to their true 
positions, at d.d' and r"V, as the highest and lowest points of 
the required shadow. 

2°. To find any intermediate points. 

Any horizontal plane, as q'D\ cuts from the surface of the 
piedouche a circle, q'D' — gsf\ and from the cylinder of rays, 
whose given base is the circle, BKF — F'A'B', a circle equal to 
the latter circle, and which is the shadow of BXF — F'A'B' on 



A 



PlXt 




"b'L«: 



F^---- - 



36 



38 




a" o ithja f I el 



PI XI 




SHADES AND SHADOWS. 89 

;he plane q f D\ The intersection of this auxiliary shadow with 
the circle qsf—q'D r , gives two points of shadow on the given 
surface (42). 

The centre of the auxiliary shadow is b\b, where the ray Ab — 
A!b\ through the centre of the upper circle of the concave sur- 
face, pierces the plane q'T)'. Then, drawing an arc, fq, with a 
centre, b, and a radius equal to A'B', we find /and q, horizontal 
projections of two points of the required shadow. Projecting 
these points on q'T)', the vertical trace of the horizontal plane 
containing them, gives q' and/ 7 for their vertical projections. 

Two points can be similarly found, on any other auxiliary 
horizontal plane between the highest and lowest points, as shown 
at e,e', on the plane e'a\ and at^,^' on the plane p'c' . 

Discussion. 

First. — The whole of the shadow just found, would be real, 
only in a geometrical sense ; or upon a hollow and transparent 
piedouche. Observing that the highest point of the curve of shade 
would be found by drawing a tangent ray, parallel to A'c"\ and 
counter- revolving as before, it is plain that the highest point of 
shadow is below the highest point of shade. Hence the shadow 
on an opaque solid piedouche will be real, from its highest 
point to its intersection with the curve of shade. The real part 
of the curve of shade was described in Prob. XXVIII. (Kem. #). 
This real part will cast a shadow on the lower part of the 
piedouche. 

Thus the complete line of separation, between the illuminated 
and the dark portions of the piedouche, consists of the real por- 
tion of the shadow of the upper circle, the real part of the curve 
of shade, and the shadow of this real shade. 

Second. — The latter shadow is thus found. See PL XL, Fig. 
36, which represents, in horizontal projection, and sufficiently for 
purposes of explanation, the shadow just mentioned, together 
with the entire shadow of the piedouche on the horizontal 
plane. 

The shadow of the curve of shade upon the lower part of the 
piedouche, is found by the special method of two auxiliary 
intersecting shadows (66-lst. Kern. a). Thus, cjkt and pmu are 
shadows of the real parts of the curve of shade, on the horizontal 
plane. They are found from the projections of those parts. 



90 GENERAL PROBLEMS. 

shown in PL X., Fig. 31, and just as any shadow is found on a 
plane of projection (30). 

The circle sr is the shadow of an equal circle, UT, near the 
foot of the piedouche ; r and s, the intersections of these sha- 
dows, are the shadows of points of shadow cast by the curve 
of shade on the circle UT. That is, rays, as sU and rT, through 
s and r, will meet both UT, and the curves of shade. U and T 
are the shadows, on UT, of the points cut from the curve of 
shade by these rays. 

Third. — pq, and ut, are the shadows of the unreal portions, 
1U2, and QrR, PL X.. Fig. 31, of the curve of shade. 

These unreal shadows join the real ones at jp, q, u, and I, 
forming cusps of the first order. 

This result being apparent, it enables us to learn, as we could 
not easily do from an inspection of the curve of shade only, in 
PL X., Fig. 31, the real nature of the tangent cylinder of rays, 
whose curve of contact with the piedouche is the curve of 
shade. 

Considering putq as its base, and that its elements are parallel, 
it is now evident, that above the point 2,2', for example, its sur- 
face curves upward, and backward towards the vertical plane ; 
and that below the same point, its surface bends downward, and 
backward towards the vertical plane ; so that it has an edge on 
the tangent ray at 2,2'. A plane through this ray, or element, 
and tangent on the same, to both the branches, just described, 
of the cylinder, is an osculatory plane to the curve of shade at 
2,2' ; and its horizontal trace is a tangent to the base of the 
cylinder at q, PL XL, Fig. 36, between hq and pq. But the 
tangent ray at 2,2', being such a salient edge as just described, a 
plane section, as the base of the cylinder, will present a cusp as 
at q, where that edge pierces the plane of the section. 

Observing, further, that the cylinder of rays, to which the 
plane of rays is tangent, is, itself, tangent to the piedouche at R, 
the plane is also tangent to the piedouche at R. Hence its hori- 
zontal trace can readily be found. 

Fourth. — The rest of the shadow, PL XL, Fig. 36, is thus 
composed: ahc is the shadow of that lower semicircle of the 
upper base of the piedouche which is towards the light, deb is 
the shadow of the opposite upper semicircle, KSB. ah and cd 
are the shadows of the elements of shade of the upper base at K 
and B. Likewise, nmy is the shadow of the upper semicircle, 



SHADES AND SHADOWS. 91 

N"HY, of the lower base of the piedouche, and ~Nn and Yy are 
the shadows of the elements of shade of the lower base, at N 
andY 



§ III. — General Problem in Review of Shades and 
Shadows, determined by Parallel Rays. 

54. No better problem can perhaps be found for this review 
than that of the shades and shadows on the Eoman Doric column, 
embracing, as it does, the shadows both of straight and of curved 
lines ; on planes, and on a variety of single curved, and double 
curved surfaces. 

Description of the Column. — PI. XT., Fig. 37, represents, with- 
out regard to precise architectural proportions, a fragment of the 
shaft and base of the column. 

Indicating its horizontal circles by their radii, the circle Ca is 
the plan of the shaft a'SK. Next, Gb is the plan of the smallest 
section of the scotia e'b'c" . Then Gc is the plan of the upper 
and middle fillets, c'g r and c"h' . And Cd is the largest section of 
the upper torus, c"d'h'g' . Ce — e'n' is the lower fillet. CF — -fn' is 
the lower torus. FGH — F'H' is a square member, called the 
plinth, a'c'g' is the foot of the shaft, and is a concave double 
curved surface. The fillets are short vertical cylinders. 

PL XL, Fig. 38, similarly represents the principal parts of the 
capital and shaft of the column. a"ahw — a'b'w'g"h' is the 
abacus, whose horizontal sections are squares. The portion, 
on — o'n'n", of the abacus, is the cyma reversa, and is cylindrical. 
Cg and C/"are the greatest and least circular sections of the half 
torus g'g'fi ', called the echinus. Gf—f'f" is a fillet. Between 
0/ and Ce is efi'f' ', the cavetto. Gc is the neck of the column, 
limited by a small torus, not shown, and called the astragal, just 
under which is the lower fillet and then a double curved concave 
surface similar to the foot of the column. 

Thus the whole column, between the plinth and the abacus, is 
a surface of revolution having a vertical axis. 



92 GENERAL PROBLEMS. 



Problem XXXI. 

To find the Shades and Shadows of the Shaft and Baseof aRomcn 

Doric Column. 

Such parts of the solution as present peculiar features, will be 
figured. Others will be merely referred to previous problems 
containing similar constructions. 

1°. The shaft is strictly a double curved surface, its upper 
diameter being, in practice, a little less than its lower, and its 
generatrix slightly curved. The fragments shown in the figures 
may, however, be treated as vertical cylinders, in finding their 
elements of shade. 

2°. The shadow of the shaft on the foot of the column, is 
found by the special method of (43) (Prob. XVI,, 3°), its hori- 
zontal projection being known as a straight line, tangent to the 
plan of the shaft. 

3°. The element of shade, and the upper circle of the fillet, cast 
shadows on the upper torus. See PI. XII., Fig. 39. 

The element of shade is the vertical line at N1SP, and its shadow 
is found as in (2°). Thus, XK is the indefinite horizontal pro- 
jection of the shadow, NX being the intersection of a vertical 
tangent plane of rays, at X, with the torus, and therefore paral- 
lel to the horizontal projections of rays, da — d'a' is any assumed 
horizontal circle of the torus, intersecting the indefinite shadow 
XK at a' whose vertical projection, a', is on d'a', the vertical 
projection of the assumed circle. 

This shadow begins at X', the foot of the element of shade. 

The upper circle casts a shadow which is found by the method 
of (42), see also (Prob. XXX— 2°). Thus, the ray, Gth—G'k', 
through the centre of the upper circle, pierces an auxiliary hori- 
zontal plane, h'e' , cutting the torus in a circle, at hft . The cir- 
cle, yb, with centre h and radius hy, equal to GrV, is the auxiliary 
shadow of the upper circle, GV, on the plane h'e'. This shadow 
meets the circle eh, cut from the torus by the plane h'e', at two 
points, one of which is b. Then b is vertically projected at b', 
on the vertical projection, e'h', of the circle cut from the torus. 

On a complete construction, the latter shadow will intersect 
the shadow of the element of shade, just before found, and the 



PL XII 




PLXII. 




L J - 














/• 








1 

J -4- 


r 


.'— 


— 


_ 


— 


&_ 


sJ 1 


/ ; i: I- i i i r 





SHADES AND SHADOWS. 93 

curve of shade of the upper torus, in points which the student 
may construct. 

4°. The curve of shade on each torus in found as in Prob. 
XXVII. 

5°. The curve of shade of the upper torus casts a shadow on 
the middle fillet, which is found as in Prob. IY. But PI. XII. , 
Fig. 39, serves, however, to show how this shadow can be found, 
having given but one projection of the curve of shade of the 
torus. Let EV be a fragment of the vertical projection of the 
curve of shade of the torus. Let o' be a point on this curve, 
which is supposed to cast a shadow on the fillet. This point is 
transferred, on a circular section of the torus, to a point whose 
revolved position is D'jD", found by revolving the semicircular 
meridian curve, in the plane DGGr', about its vertical diameter 
at S. 

In the counter revolution about S, D'D" proceeds to D, in 
horizontal projection ; and in the counterrevolution about the 
vertical axis at G, it returns to o, the desired horizontal projec- 
tion of o' '. 

This done, the shadow of o,o' is found by the simple direct 
method of (40). Thus, op, the horizontal projection of the ray 
through o,c/, pierces the fillet at a point of shadow whose hori- 
zontal projection is p. The vertical projection, p\ of this point, 
is at the intersection of the projecting line, pp', with the vertical 
projection, o'p' , of the same ray. 

6°. Part of the same curve of shade on the upper torus, casts 
a shadow on the scotia. PI. XII., Fig. 40. This curve of shade 
being of double curvature, its shadow on an auxiliary plane, P 
as in the special method of (42), will be an irregular curve, a num- 
ber of whose points must be found, and joined together, to give 
the auxiliary shadow, whose intersection with the circle cut from 
the scotia by the plane P, would be a point of the required 
shadow. 

Hence the special method of (42), which, as seen in Prob 
XXX., is so convenient when the shadow is cast by a circle, is 
no more convenient than the direct method (82) when the curve 
casting the shadow is of double curvature, as in this case. 

For this reason the direct method (82) as applied in problems 
like the present, is here used. Let h'c' be a fragment of the ver- 
tical projection of the curve of shade of the torus. Let V be 
any point on this curve, whose shadow on the scotia is to be 



94 GENERAL PROBLEMS. 

found. It is transferred, on a circular section of the torus, to a\ 
whose horizontal projection is a. In counterrevolution about a 
vertical axis at D ; a\a returns to b', b. Now, b'r' is the vertical 
trace of a plane of rajs, taken through b,b r , and perpendicular 
to the vertical plane of projection. It cuts from, the scotia the 
curve efr — e'f'r' ; which is found, as fully shown, by auxiliary 
horizontal planes, as Q>'f. This plane cuts from the scotia the 
circle Q'f — C/J and from the curve c'r' the point/ 7 , whose hori- 
zontal projection, f is on the horizontal projection, Of of the 
circle containing it. Having: thus constructed the intersection 
of the plane of rays with the scotia, by means of a system of 
auxiliary planes, draw the ray bs — 6V, which pierces the curve 
efr — eV at s,s', the required shadow of b,b'. Other points of 
shadow on the scotia may be similarly found. 

The plane of rays may also be taken vertical. 

7°. There will be two points of the curve of shade b'c\ the 
rays through which will pass through the points, as e,e', on the 
lower edge of the fillet. That part of the curve b'c\ lying 
between these points, casts a shadow on the fillet. These points 
themselves cast shadows on the lower edge of the fillet, and the 
part of this edge between these points, will cast a shadow on the 
scotia. The latter shadow will be found as in Prob. XXX. 

8°. The curve of shade of the scotia will be found as in Prob. 
XXVIII. 

9°. IfJ on account of a certain direction of the light, the 
curve of shade of the upper torus casts a shadow on the lower 
torus, it will be indicated by the passing of some of the rays, 
similar to 5s, in front of r,r\ so as not to cut the corresponding 
curve similar to efr. This shadow will be found as in (6°). 

10°. The shadows of the lower fillet on the lower torus are 
found as in (3°). 

11°. After finding the curve of shade on the lower torus as in 
Prob. XXVII. , its shadow on the top of the plinth will be found, 
a point at a time, as in Prob. III. 



SHADES AND SHADOWS. 95 



Problem XXXII. 

To find the Shades and Shadows of the Capital and Shaft of a Roman 

Doric Column. 

In making the solution of this problem, we have — 

1°. The shadow of the upper member of the abacus on the 
cyma reversa ; the element of shade on the latter ; its shadow 
on itself, and on the lower part of the abacus. These shadows 
form one topic, being all found in the same general way, as seen 
in PI. XIL, Fig. 41. 

The cyma reversa, being bounded by cylindrical surfaces, it is 
necessary to have an auxiliary elevation of it, as it appears when 
seen in the direction of the arrow. The point T" then is the 
auxiliary vertical projection of the lower front edge, NY — T'A'. 
Seen in the direction of the arrow, the point, B,B', of the ray 
AB — A'B r , appears at a distance, A!s, below T" — T'A', and at 
a distance, rB, to the left of NY — r" . Therefore, make T'V 
equal to AY, and r''B" equal to rB, then T"B" will be the 
auxiliary vertical projection of a ray, or the trace of a plane of 
rays through NY — T'A' — T". This plane cuts from the front 
of the cyma reversa the line u" — uu\ the upper shadow on the 
cyma reversa. A parallel, and tangent, plane of rays, La", 
gives the element of shade, n" — nn\ and its shadow, a" — aa f , 
Another plane of rays, through the edge below a" — aa', would 
give the shadow on the lower part of the abacus, not shown. 

2°. The shadow of the edge h'g", PL XL, Fig. 38, on all the 
cylindrical parts below, is found as in Prob. IV. 

The shadow of the edge which is perpendicular to the paper 
at h\ on all parts below, is bounded by the intersection of a 
plane of rays through this edge with those lower surfaces. This 
intersection will appear in vertical projection as the vertical pro- 
jection of a ray at h' (43). 

3°. The shadow of the lower front edge of the abacus, ET — 
E'T', Pi. XIL, Fig. 42, on the echinus is found thus : Let N'K' 
— Ne</ be a circle, cut from the echinus by a horizontal plane, 
N'K'. A ray, cd — c'd\ through any point, as c,c\ of the edge 
casting the shadow, will meet this plane in a\a", which determines 
dg, the shadow of ET— E'T' on the parallel plane, N'K' (5). 



96 GENERAL PROBLEMS. 

This auxiliary shadow meets the circle, 1$eg — WK.', at e,e' and 
g,g', two points of the required shadow, which is thus found by 
(42)— see also Prob. XXX. 

Observing that un is equal to Stf, the highest point, /', of this 
shadow, may be thus found. The plane of rays, through RT — 
R'T', intersects the meridian plane, Dn : in a line, nb, which is the 
projection of the light on the plane Dn. The intersection of 
this trace nb with the echinus is the point f. Revolve bn to ba", 
whose vertical projection is a'b' . Draw RT/, parallel to a'b', 
and revolve L to /', and f is the required highest point of the 
shadow. By drawing a ray through /', we could find the point 
whose shadow is f . 

The shadow of RT — R ; T ; is real, above its intersection with 
the curve of shade of the echinus, which is found as in Prob. 
XXVII. 

The remaining shadows involve no new operations. 



SHADES AND SHADOWS. 97 



SERIES II. 

SHADOWS DETERMINED BY DIVERGING RAY3. 

SECTION L 
General ^Principles. 

85. Having seen, in the preceding general problems, how 
shades and shadows are found, when the rays of light are parallel, 
it now remains to examine Arts. (2) and (16) somewhat in detail. 
When a bodv, B, is illuminated by a single luminous point, that 
point may be considered as the vertex of a cone of rays, C, tan- 
gent to the body, B. The line of contact of C and B, is the line 
of shade of B. The interior of the cone, beyond B, is the 
shadow, in space, of B, and that portion of any secant surface, 
S, which is within the cone, and beyond B, is the shadow of B 
on S. If, now, B be illuminated simultaneously by two points, 
P and P r , a portion, both of B and S, will receive light from 
neither point. This is the total shade on B, or shadow on S, 
respectively. Another portion of B and of S will receive light 
from both points. This will be their completely illumined por- 
tion. A third portion of B, and of S, will receive light from 
one or the other of the points, P and P', but not from both. 
This is their penumbra, or partial shade, or shadow. 

86. Let this case be extended to an assemblage of luminous 
points, forming a luminous body, L. In general, L and B will 
be of different sizes, hence they may have two common tangent 
cones of rays inclosing them ; one with its vertex beyond both 
bodies, the other with its vertex between the two bodies. The 
area, or zone, on B, between the curves of contact of the two 
cones, will be the partial shade of B. The annular space on S, 
between the intersections of the two cones with S, will be the 
penumbra or partial shadow of B on S — the part which is reached 
by only a part of the rays from L. 

7 



98 GENERAL PROBLEMS. 

87. See, now, PI. XIII. . Fig. 4A. where AC represents a 
luminous body ; BD, an opaque body, and PQ an opaque sur- 
face which receives the shadow of BD. The dark space, bd, is 
the total shadow, and is bounded by the cone of rays, tangent to 
both bodies, and having its vertex on the same side of both. 
The lighter space, ca, is the penumbra, and is bounded by :_t 
tangent cone whose vertex is L. 

88. Regarding L at first as the original source of light, and the 
vertex of the single cone Lea, it may then be regarded as the 
vertex merely of a complete cone of two nappes, in the outer 
nappe of which is inscribed the extended source of light, AC, 
around which, and the body BD, may be inscribed the second 
cone which determines the shadow bd. 

89. It should be noted, that the volumes of rays will only be 
cones, when the bodies, AC or BD. are similar, or have similar 
curves of contact ; or when, at least, they can both be inscribed 
in one given cone, so as to have a continuous curve of contact 
with that cone. 

In other cases, the volumes of rays will be bounded by warped 
surfaces, but these cases are unimportant, and will not be further 
considered. In every case whatever, the surface of rays will be 
a ruled surface, since ravs of light are straight lines. 

90. For the practical case, let L be the sun, and B a body 
near the earth, S. Here the angular breadth of the penumbra 
of shade upon a spherical body, B, would be, as found by a 
simple calculation, only yfy of the radius of B. 

See PI. XIII. , Fig. 43, where is the sun's centre, a, the 
centre of a small spherical body near the earth, and Y, and v, 
the vertices of the two tangent cones of rays already described. 
Then, on account of the relatively great distance of the sun, 
AB, in the triangle ABC, becomes sensibly equal to his diame- 
ter, and aC and ab, radii of the small body, being perpendicular 
to BC and AC, respectively, the chord bC sensibly reduces to 
the segment, bC, of the side AC, and ABC and abC are similar 
triangles, and AB : BC : : bC : aC : or 

885 000 : 95 000 000::'&C : aC ; or 
bC = aC very nearly. 
107 

This penumDra of shade, whose breadth is bC, may therefore 
be disregarded on terrestrial objects, and the solar rays may, 
accordingly, be considered parallel- 



SHADES AND SHADOWS. 9£ 

Remark.- -The three following problems, embracing both 
shades and shadows upon a variety of surfaces, plane, single- 
carved, and double-curved, are meant to be sufficient to initiate 
the student in the solution of problems in which the source of 
light is supposed to be a near luminous point. 

After tie explanations already made, separate statements of 
"principles" will not precede the " constructions " of these 
problems. 

SECTION n. 

Problems, involving divergent Rays. 

Pkoblem XXXIII. 

To find the shadow of a semi-cylindrical abacus, upon a vertical 

plane through its axis, 

PL XIII., Fig. 45, FcE— FVE' is the half abacus, with hori- 
zontal semi-circular bases, whose diameters are in the vertical 
plane of projection. 

L,I/ is the source of light. A line from this point to any 
point of an edge, or element, of shade of the abacus, is a ray of 
light, whose intersection with the vertical plane of projection 
will be a point of the required shadow. 

LD is the horizontal trace of a vertical tangent plane of rays, 
which determines the element of shade d — d'd". Then from 
d,d" to E,E', is the edge of shade of the upper base, and from 
d,d' to F,F', is the edge of shade of the -lower base. 

These edges, and the element of shade, cast the line of shadow 
which is required. 

The shadow then begins at FjF'. Any point, as a, a', casts a 
point of shadow A, A', which is found where the ray La — LV 
pierces the vertical plane of projection. 

The element of shade, d — d'd", casts the shadow D'D", and 
the curve D"E' is the shadow of dE—d"W. The shadow of 
c,c\ the foremost point of the lower edge of shade, is C, the 
lowest point of the shadow ; which is now fully determined. 

The shadow of a tangent at c,c' will be a tangent at 0', parallel 
to the ground line. Other obvious tangents, useful in sketching 
the curve, can readily be determined. 



100 DIVERGENT RAYS. 



Problem XXXIY. 

To find the curve of shade on a sphere, the light proceeding from an 

adjacent point. 

PI. XTIL, Fig. 46. Let the centre of the sphere, Sa'dc, be in 
the ground line, and let L,L' be the luminous point. 

1°. To find the highest and lowest points of shade. — LO is the 
horizontal trace of a vertical plane of rays containing these 
points. After revolving this plane about the vertical diameter 
of the sphere, till it coincides with the vertical plane of projec- 
tion, the luminous point will appear at I/',!/", and the great 
circle cut from the sphere, at Sa'dc. The revolved rays, \I"h'", 
\J"l"\ tangent to this circle, then determine h'" ,h" , and U", the 
revolved positions of the required points. In the counter revo- 
lution, h'"fi" proceeds in the horizontal arc, h"h — h'"h' to its 
true position h,h\ The construction of 1,1', the lowest point, 
will be given presently. 

2°. To find the foremost and hindmost points. — Gro through a 
series of operations, similar to the foregoing, upon a plane of 
rays, I/O, perpendicular to the vertical plane of projection, 
beginning by revolving it into the horizontal plane of projection, 
about the horizontal diameter, whose vertical projection is 0. 
This will give the points/,/ 7 , and e,e'. The middle point, n, of 
/V, is the vertical projection of the centre of the circle of shade, 
then l\ the vertical projection of the lowest point, is at the inter- 
section of l'"l\ the vertical projection of an arc of counter revo- 
lution, with the diameter, h'n'V. Then V is horizontally projected 
in LO, at I. 

3°. The points on the circles which are in the planes of projection, 
are found by drawing tangent planes of rays, perpendicular to 
the planes of projection. As the centre of the sphere is in the 
ground line, one circle of the diagram represents both of its pro- 
jections. Then LV and \Jb' are the vertical traces of tangent 
planes of rays perpendicular to the vertical plane. They give 
a', a and b\b, as the points of shade, on the circle, Sb — Sa dc, 
which is in che vertical plane of projection. 

Likewise, Lc and Lc?, the horizontal traces of vertical tangent 
planes of rays, give the points c,c ? and d,d f on the circle Sa'dc— 
S5, which is in the horizontal plane of projection. 



pixm 




n xiii 




SHADES AND SHADOWS. ' 101 

The construction of other points, as, for instance, those on the 
great circle, perpendicular to the ground line, is left for the 
student. 

Having now eight points of shade, the curve of shade may 
be sketched. To avoid confusion, only the vertical projection 
of the shade is shown, the horizontal projection of the points 
of the curve of shade being left unconnected. Also, the whole 
shade on the front hemisphere, is represented as visible, for the 
sake of clearness, though only a quadrant of the sphere is in the 
first angle. 

Problem XXXV. 

Having a niche, whose base is produced, forming a full circle ; and 
a right cone, the centre of whose circular base coincides with the 
centre of the base of the niche, it is required to find the shades and 
shadows of this system, when illumined by an adjacent point. 

1°. The shadows on the base and cylindrical part. 

PI. XIIL, Fig. 47. The given magnitudes being familiar 
ones, their projections, in the position described, may be under- 
stood from the figure. L,I/ is the luminous point. 

La is the horizontal trace of a vertical plane of rays, through 
the edge A — A 7 A" of the niche, and limits the shadow, Aa, of 
the niche upon its base. From a,a" , the shadow of the same 
edge is the element, a — d'a' , of the cylindrical part of the niche, 
limited by the ray La — L'a". 

The shadow, e,e' , of a point, F,F ; , of the front circle of the 
spherical part, upon the cylindrical part, is seen to be found as 
in Prob. XVILL, the ray being drawn through L,L'. 

2°. The shadow of the spherical part on itself, is found essentially 
as in Prob. XXIX., only each point of this shadow requires a 
separate auxiliary plane, whose vertical trace will pass through 
L'. Thus, L'N' is the trace, on the front face of the niche, of an 
auxiliary plane perpendicular to the vertical plane of projection. 
The ray, in this plane, meets the semicircle, cut by it from the 
niche, in a point of shadow. Wl"W, described on M'N' as a 
diameter, is the semicircle just named, after revolution about 
M'N 7 into the front face of the niche. The point L,L' is at a 
distance, LJ, from the front of the niche, hence, making L'L" 
equal to LJ, and perpendicular to L'N', it results that L // M7 /< 
is the revolved position of a ray, giving l" for the revolvec 



102 DIVERGENT RAYS. 

position of a point of shadow. By counter-revolution about 
L'N', the point I" returns to I' , its true position as the shadow 
ofM'. 

3°. The method of Prob. XXIX. (2c? solution) is here shown, 
also. Thus, let DE be the horizontal trace of a vertical plane, 
parallel to the front of the niche. It cuts from the spherical 
part, the semicircle DE— D/'E'. The raj, LY— I/O', through 
the centre of the front semicircle, AYB, pierces this plane at o,o'. 
The horizontal line, o'a'" , of this plane, limited, at a"' ' , by the 
ray L'A'V", is the shadow of the radius O'A" on. the plane DE. 
Hence d"f, with centre o', and radius, o'a'", is an arc of the 
shadow of the front circle, on DE, and /', its intersection with 
the semicircle D/'E', cut from the niche by the plane DE, is a 
point of shadow on the spherical part. 

It will now be easy to find the point n' ; the point of which 
/' is the shadow ; and the horizontal projection of the curve 
T'l'n', remembering that it is a plane curve (83). 

4°. To find the elements of shade of the cone, and its shadows. 
Draw the ray LY — LY' and find U, where it pierces the plane 
of the base of the niche. JJt and JJq are the horizontal traces 
of planes of rays, tangent to the cone, and tV — t'Y', and qY — q' 
Y', are its elements of shade. 

The traces JJt and ~Uq bound the shadow on the base of the 
niche. At m,m! and p i p / the shadow on the cylindrical part of 
the niche begins. The ray LY — L'Y' pierces this part at v,v\ the 
shadow of the vertex on the interior of the niche. The shadows 
of the elements of shade being curves, other points besides m,m' 
and p,p' must be found. To avoid the acute intersections of pro- 
jecting lines with q'Y' and t'Y', find, by division, their middle 
points, giving g,g' and h, h'. Kays, L# — Jj'g' and LA — L'A ', through 
these points, pierce the cylinder of the niche, produced, at s,s' 
and r,r' , respectively. Through these, and the previously found 
points, the indefinite shadows, v'm's' and v'p'r', of the elements of 
shade, can be drawn. The points, as u, where the horizontal pro- 
jections of the rays just drawn meet the horizontal traces of the 
planes of shade, are the points in which those rays pierce the hori- 
zontal plane, and should therefore appear in perpendiculars to the 
ground line through u', etc., which are their vertical projections. 

The niche conceals the horizontal projections of the shades 
and shadows except the portion of shade YtK. on the cone, and 
tJcK, a portion of the shadow of the cone, on the base of the niche 



SHADES AND SHADOWS. 103 



PART II 
SHADES AND SHADOWS IN ISOMETRICAL PROJECTION. 

SECTION I. 
General Principles. 

91. Isometrical projections of shadows may be found by two 
quite different methods. 

First. — They may be found directly on the isometrical projec- 
tion of the object which receives them. Second. — They may be 
first constructed in ordinary orthographic projection, and then, 
from such projections, their isometrical projections may be found. 
Examples of both of these modes of procedure will be given 
among the following problems. 

92. Since the essential feature of isometrical drawing is, that 
it shows three dimensions, at right angles to each other, in their 
real size, it follows, that this kind of drawing is chiefly useful in 
its applications to rectangular bodies. This fact, also, will be 
illustrated in the following constructions. 

93. In isometrical projection, only one principal plane of pro- 
jection is used ; but, according to first principles, two projections 
of a line, i. e., its position with respect to at least two planes, are 
necessary to determine its position in space ; hence, in isometrical 
projection, the rays of light are more conveniently determined by 
referring them to isometrical surfaces, as those of the body illu- 
minated, than to planes of projection. Hence, again, isometrical 
drawing is most useful in connection with rectangular objects ; 
since, in representing other objects, auxiliary projections of rays 
must be employed, as will soon be seen. 

In the following problems only parallel rays are employed. 
In PI. XIV., Fig. 48, the light is supposed to enter the cube at 
the upper left hand corner, E, and to leave it at the lower right 
hand corner, J, and this is the conventional direction of the light 
generally, in isometrical drawing. 



104 ISOMETRICAL SHADOWS. 

The raj, being thus a diagonal of the cube, makes an angle of 
35° — 16' with each of its faces. Its projections, as EK and EI, 
on those faces, are diagonals of the faces, and therefore make 
angles of 45° with the edges of the cube. In the absence of 
these planes and edges, as when a curved surface is the subject 
of a problem, it may be convenient to know the angle made by 
the ray with the plane of projection. This we next turn to find. 



Problem XXXYI. 

To find the angle made by the isometrical ray of light with the 
isometrical plane of projection. 

The point, C, Fig. 48, is the projection of the diagonal from 
the foremost to the hindmost points of the cube, hence the plane 
of projection is perpeDdicular to this diagonal. Then, in Fig. 49, 
let ACBD be the plan of a cube, and D'C'E'Gr' its elevation; 
the cube being in the fourth angle, with its vertical faces making 
equal angles with the vertical plane of projection, parallel to the 
paper. Then PQ, perpendicular to the diagonal DT', is the 
trace, on the plane of the paper, of the plane of isometrical pro- 
jection, and AB and B'E 7 are the two projections of the ray seen 
at EJ, in Fig. 48. Observe that EK, EI, and KI, Fig. 48, are 
parallel to the isometrical plane of projection. Then B'G', Fig. 
49, is the trace of a plane, parallel to the isometrical plane PQ, 
and the point A,B' is its own projection on this plane, and the 
point B,E' is projected upon it by the perpendicular E'^, which 
is seen in its true size. But the diagonals of the cube being 
equal, D'F'^ the line AB— B'E', also Wg = § of D'F, and hence 
of the ray AB — B'E', and if then the ray, considered as radius, 
be called 1, Wg = ^ will represent the sine of the angle made by 
the ray with the parallel plane B'Gr' and hence with the isome- 
trical plane PQ. But | is the Xat. sine of 19° — 28', very nearly, 
which is therefore the desired value of the angle made by a ray 
of light with, the plane of isometrical projection. 



SHADES AND SHADOWS. 105 

SECTION II. 
Shades and. SliacLo^ws on Isometrical IPlanes. 

94. Shadows on isometrical planes will be the intersections of 
rajs, through points casting shadows, with their projections on 
such planes, or with the traces, on the same planes, of any planes 
of rays containing those points. 

Problem XXXYII. 

Having given a cube, with thin plates projecting vertically and for- 
ward, in the plane of its left-hand bach face, to find the shadows 
of the edges of these plates upon the cube and its base. 

PL XIV., Fig. 48. To find the shadow of KB upon the top of 
the cube, to which it is parallel. — A a and Bb represent the rays 
themselves, through A and B ; hence, as AE and Be are project- 
ing lines, perpendicular to the top of the cube, Ea and cb, on, 
and parallel to EK, are the projections of these rays, upon the 
face ECKL. These rays meet their projections at a and b, the 
shadows of A and B, hence ab is the shadow of AB. ~Eabc is 
now evidently the shadow of the rectangle ABcE. 

To find the shadow c/EDFG. — Here BE, being perpendicular to 
the face ECIGr, its shadow on that face is in the trace, ET, of a 
plane of rays, through DE, upon that face, and is limited at d, by 
the ray Dd. From d, dh, parallel to DF, is the shadow of the 
portion, DH, of the edge DF. Then h¥ is the shadow of HF 
on the plane of the lower base of the cube. 

We see that PL XIV., Fig. 50, shows the case in which the 
line, AB, casting the shadow, coincides, in projection, with a ray. 
Either point, as b, of the shadow, may be found, either by pass- 
ing a plane of rays through B, and perpendicular to the top sur- 
face FHK, or to the face FKI. The line BD, perpendicular, 
and the trace Db parallel to FI, determine a plane in the former 
position, and b, the intersection of the trace Db, and ray Bb, is 
the shadow of B. BE, parallel to KH, and the trace E5 in the 
direction of EI, Fig. 48, determine a plane in the latter position, 
above named ; and b is seen to be, as before, the shadow of B 
on FKI. 



106 ISOMETRICAL SHADOWS. 

Observe that E&, and EI, Fig. 48, make angles of 60° with a 
horizontal line. 



SECTION IIL 
Shades and. Shadows on Non-Isometrical X^lanes. 

Problem XXXYIII. 

To find the shadow of a hexagonal cupola on a coupled roof, ont 
face of the cupola making equal angles with two adjacent walls of 
the house. 

PL XIV., Fig. 51. To locate the walls and roof Let A. f "A!'Y 
and YA"n' be portions of the adjacent walls, which are at right 
angles to each other. 

Suppose the inclination of the roof to the horizontal plane, 
A!" 'A'W, of the eaves, to be 30°. Find 0, the middle point of 
A'" A". Then make OA', horizontal, and equal to OA'", and 
make OA'B'— 30°, then B', the intersection of the vertical OB', 
with A'B', is the extremity of the summit of the roof; through 
it B'A'", B'A", and B'M, may be drawn, and through M, the 
ends of the roof, parallel to A'"B' and A^B'. 

1°. To locate the cupola. — Let n be the point at which its axis, 
riNj pierces the centre line, On, of the plane of the eaves, and 
let nv be the half width of the square vgde, in which the circum- 
scribing circle, ache, of the base of the cupola is inscribed. The 
semicircle. akh, represents the semicircle, acb, after revolution 
about the axis ah, till it is parallel to the plane of the paper, i. e., 
to the isometrical plane of projection. Hence trisect akh, and 
at k and h draw kc and hd, the projections of the arcs of coun- 
ter-revolution described by k and h; and ac, cd, and db, will be 
three sides of the base of the cupola. The remaining sides are 
parallel to these three. At the corners a, c, etc., of the base, 
draw the equal vertical edges aC, cl, etc., and join their upper 
extremities, which will complete the cupola. 

2°. To construct the intersection of the cupola with the roof. — The 
plane of the face led gives the trace, um, on the plane of the 
eaves, and the trace u\J, on the vertical plane through the ridge 
B'M. Then Um is its trace on the front roof, and CD, the defi- 
nite intersection of the face led with the front roof. Similar 



SHADES AND SHADOWS. 107 

planes, through lib. and Ga, give the traces N?i, on the front 
roof, and NP, on the back roof, and give B and A, as the inter- 
sections of these edges with those roofs. 

The vertical plane through the ridge, cuts the edges, ac and 
be, of the base, at s and t. Yertical lines, sS, and tT, from these 
points, meet the ridge at S and T, where it meets the walls of 
the cupola. From S and T, draw SA and SO ; TB and TE, 
which, with AF and BD, complete the required intersection. 

3°. To find the shades and shadows. — The auxiliary planes, just 
used, being planes of rajs, determine the faces whose upper 
edges are IJ, JH, HK, and KL, as in the shade. Kays, as It 
and Hjo, through the extremities of these edges, meet the traces, 
Cr and Bp, of the planes of rays on the roof, produced, at r, p, 
etc. Then Drq'poQE is the required shadow, of which only 
the part on the actual roof is real. 

The student may reconstruct the figure, with the face ID 
parallel to the wall YA"n. 



SECTION IV. 
Shades and. Sh.ad.ows on Single Curved Surfaces. 

Pkoblem XXXIX. 

To find the elements of shade on an inverted hollow right cone, the 
shadow on its interior, and the shadow of one of its elements of 
shade on an oblique plane. 

1°. To find the elements of shade. PL XIV., Fig. 52.— Let the 
isometrical ellipse, SQTK, be the upper base, OV the axis, and 
the tangents, VQ and VN", the extreme elements of the cone. 
VOR is a plane of rays, and its trace, OR, on the base, meets 
the ray VR, parallel to EJ, Fig. 48, at R, the intersection of a 
ray through the vertex, with the plane of the base. Then RS 
and RT are the traces, on the plane of the base, of tangent planes 
of rays, which determine TV and SV as the elements of shade. 

2°. To find the shadow on the interior. — ~R.b is the trace of a 
secant plane of rays, which cuts from the base the point a, cast- 
ing a shadow, and from the cone, the element, bV, receiving the 
shadow. Then the ray, a A, determines A, the shadow of a, 



108 ISOMETEICAL SHADOWS. 

In the same way, other points, C, etc., may be found. The 
shadow ends at S and T. The shadow of e is the lowest point 
of shadow, e being the point furthest from the element which 
receives its shadow. The shadow of e is E, on the element fY 
(undistinguishable in the figure from NY). The shadow on 
the element NY, is found by drawing a trace BN. 

3°. To find the shadow of the front element of shade on an oblique 
plane, BLFL — DKFI is in an assumed horizontal plane through 
FI ; and as its position, relative to the cone, is undetermined by 
the projection, it is, when produced, assumed as cutting the axis 
OY at m. Then make Tt equal to Om, and t is the projection 
of T on the plane DKFI. Hence tu is the trace, on this plane, 
of a vertical plane of rays through T ; uU is its trace on the 
vertical surface BDF, and Uy, on the oblique plane. The ray, 
Ty, therefore gives y, as the shadow of T. 

Another point of the shadow of TY, is where it pierces the 
oblique plane. Xow tm, parallel by construction to TO, is the 
trace of a meridian plane, OTtm, on the horizontal plane DFI. 
This meridian plane being vertical, it intersects the vertical sur- 
faces, BDF and BDK, in the vertical lines AH and gGr, giving 
GH for its trace on the oblique plane. But this meridian plane 
contains, by construction, the element of shade, TY. Hence r, 
where TY meets the trace G-Hr, is the intersection of TY with 
the oblique plane, produced, and is therefore a point of its sha- 
dow on that plane, Hence ry is the required shadow of TY. 

The shadow of SY may be similarly found. 

The distance equal to Om, might first be assumed as at^P, 
then Pm, parallel to_pO, will locate m t and t will be the intersec- 
tion of Tt with Ft, parallel to pT. 



SECTION Y. 
Sh.ad.es and. Shado^ws on Doxible-CTarved Surfaces. 

Peoblem XL. 

To find the curve of shade on a sphere.. 

The projection of the sphere on the isometrical plane of pro- 
jection is the circle CBDA, PI. XIY., Fig 53. Make the angle 
KF'E' equal to PF'C, Fig. 49, then KF will be the intersec 



PI XIV. 




I'l XIV 




SHADES AND SHADOWS. 109 

tion of the plane of the paper, which is the isometrical plane of 
projection, with an auxiliary plane, taken as a vertical plane of 
projectioD, and on which E'F' is the trace of a plane of rays. 
Project back O at 0', and make O'E'rrO'F', and each equal to 
OA. Then E / F / is the auxiliary projection of that great circle 
which is contained in a plane of rays. Project E' and F' at E 
and F ; then the ellipse on AB and EF as axes, will be the iso- 
metrical projection of the great circle just named. Planes of 
rays, parallel to E'F', will cut parallel small circles from the 
sphere, whose diameters will be chords of the horizontal great 
circle, whose isometrical projection is AHBG-. 

As these parallel circles will be projected in similar ellipses 
assume gh, kn, etc., as their diameters, and draw As and wr parallel, 
to BE, which will give the conjugate axes, es and fr, of these 
ellipses. Constructing these ellipses, and drawing rays tangent 
to them, gives points of shade, d, c, b, etc. 

The tangent at a, parallel to the rays, is the trace of a plane 
of rays perpendicular to the plane of isometrical projection, and 
gives a as the point of shade on the great circle, parallel to the 
paper, which forms the isometrical projection of the sphere. 

H and Gr are points of shade, and a line aOp is the trace of the 
plane of shade on a meridian plane parallel to the paper, giving 
p for another point of shade. Other points can be found as first 
described. 

Remark. — For an approximate construction, circles with radii, 
eh, etc., may be employed instead of ellipses, in ordinary con- 
structions. 



Problem XLL 
To construct the curve of shade on a torus. 

PI. XIV., Fig. 54. KT is the intersection of the isometrical 
plane, or plane of the paper, with the auxiliary vertical plane. 
FI/ is the ground line of the latter plane, on that auxiliary plane 
which is used as a horizontal plane of projection, and is found 
by making the angle L'FT^E'F'Q, Fig. 49. 

Tida" and a!e'h"u' are the auxiliary projections of the torus. 
The auxiliary projections of the light, in the position which it is 
desired to have on the isometrical projection, are 0"L — K'L'. 



110 ISOMETKICAL SHADOWS. 

We now, according te (91 Second), construct the auxiliary pro- 
jections of the curve of shade, and then the isometrical projection 
of the torus and its shade. 

1°. To find four 'points of the curve of shade. — From Fig. 43, it 
appears that the light makes an angle of 35° — 16' with the plane 
of the base of the cube, i. e., with the horizontal plane of projec- 
tion. Then, knowing that the horizontal plane (so called in 
Fig. 54) and seen edgewise before revolution at FI/, makes this 
same angle with the horizon, it follows that 0'''L" — RT/" is a 
ray, after being revolved about the (relatively) vertical axis 
0" — I/O', till parallel to the auxiliary vertical plane ; and RT/" 
is found horizontal. Then draw rays, tangent at e' and u\ and 
parallel at RT/", and e' and u' will be the revolved positions of 
the highest and lowest points of the curve of shade, whose true 
positions are hf and R'. 

The points of shade on the greatest horizontal section of the 
torus, are a",a r and V'f)'. 

2°. To find intermediate points. — Let the points in the meridian, 
plane, cd, be found. According to (80) 0"l and R7' represent 
the ray 0"L — RT/ after being projected on this plane, cd. 
Then 0"l" and RT" are the revolved positions, in the parallel 
meridian plane, a"b", of the projected ray. Next, draw from r, 
the centre of the semicircular part of the meridian curve, rd!" , 
perpendicular to RT", then d'",d" is the revolved position, and 
d,d' the primitive position of a point of shade ; O"— \Jh! being 
the axis of revolution. For variet} 7- of construction, revolve Yc 
to the position Yc /; , parallel to the vertical plane of projection. 
This semicircle will then be vertically projected, with the radius 
p'm! ; and by drawing p'c'", perpendicular to RT", we find the 
point of tangencj, c"' \c", of a revolved iay, parallel to RT", and 
from this, the same point in its true position, c,c f . 

3°. To construct the isometrical projection of the torus, and 
of its shade. Assume eu for the trace of the meridian plane, 
a"V\ upon the isometrical plane. 

The highest points, e and w, are projected from e' and u' upon 
this trace. The right and left hand points, E and G, are pro- 
jected from 0', and EGr is the projection of the diameter E"Gr". 
To find intermediate points, as those in the plane cd, project 
the direction of vision, sO" — s'O', upon this plane as at SO" — 
S'O'. Then revolve this projected ray, together with the meridian 
plane, about a vertical axis at 0" till parallel to the vertical 



SHADES AND SHADOWS. Ill 

plane of projection, and then by (80) the points of contact of 
projecting lines, parallel to s'"Q\ will be points of apparent 
contour in the isometrical projection. These points are best 
found by drawing radii, as m"\ perpendicular to s'"§' , which 
gives n r "n", and after counter-revolution, n,n r . There being 
evidently four such points, make b7$=ns, and then making 
Oa=Ob, make al, aB, and 6 A, each equal to bN. The oval figure 
passing through the points now found, will be the isometrical 
projection of the torus. 

For the curve of shade ; h' is projected at H, £H being equal 
to hli" . Oq — Ot, and then, qR=tK. The points c' and d' are 
then projected at C and D, at distances from eu equal to the 
distance of d from a"b". Finally, a' and b' are projected in the 
meridian curve, a"V\ whose isometrical projection is eu, at a ana 
b. Through the six points now found, with f and Jc, the points 
of tangency of tangent planes of rays perpendicular to the isome- 
trical plane, the curve of shade can be drawn. 

H, the highest point, determines the visible portion of the 
curve of shade to be^HD^. 

Remarks. — a. After all the labor of making the foregoing 
construction, it is now more fully evident, according to (92), that 
while this, and the three preceding constructions, afford good 
examples for study, they, and the present problem particularly, 
show that isometrical projection has little or no advantage in 
respect to clearness of representation, except as applied to plane- 
sided bodies having solid right angles, whose sides are equally 
inclined to the plane of projection. 

b. On account of the superior pictorial character of isometrical 
projections, and still more, of the oblique projections explained 
in a previous volume (Elementary Projection Drawing) under 
the name of " Cabinet Projections," it is of less importance to 
represent shades and shadows upon them. The cabinet projec- 
tions of shades and shadows can readily be made from their com- 
mon projections 



112 BRILLIANT POINTS. 



Book 2, 

THE FINISHED EXECUTION OF SHADES AND 

SHADOWS. 

CHAPTER I. 

THEORY AND CONSTRUCTION OF BRH.LIANT POINTS 5 
AND OF GRADATIONS OF SHADE. 

SECTION L 
Preliminary General Principles. 

§ 1°. — Geometrical Conditions for the adequate graphical repre 

sentation of Form. 

95. The obviously essential geometrical feature of a surface is 
its continuity. But the bounding surface of a volume is repre- 
sented, geometrically, as seen from a given point, by its apparent 
contour • which is only that line of the surface which is its 
visible boundary, as seen from that single point. 

Geometrically, therefore, an infinite number of consecutive 
contours, seen from as many points of view, on each of the two 
opposite sides of a body, would be necessary to completely 
represent its continuity and conformation. 

But the method of projections usually gives but two apparent 
contours ; viz., those which constitute the two projections of a 
body. Furthermore, it would be graphically impossible to 
represent consecutive apparent contours ; hence, purely geo- 
metrical diagrams, though enabling us to represent any required 
point of a surface, can never, of themselves, adequately repre- 
sent the continuity of the inclosing surface of a body. 

96. It is therefore our previous and definite conception of 
the surfaces to be represented, which enables the projections of 



SHADES AND SHADOWS. 113 

surfaces practically to express the forms of those surfaces intelligi- 
bly. This may be illustrated by the difficulty usually experi- 
enced at first, in comprehending projections of new surfaces, as 
warped surfaces, before acquiring, from models or other sources, 
some idea of their form. On the other hand, the circular and 
rectangular projections of the familiar cylinder of revolution, 
perfectly represent it to the mind, because we so well know that 
all its right sections are equal circles, and all its meridian sections 
are equal rectangles. 

97. By calling in physical considerations to the aid of geo- 
metrical ones, and confining ourselves to the sense of sight as a 
means of judging of the configuration of bodies, the action of 
light upon their surfaces would be observed at once. This leads 
to the following considerations : 

§ 2°. — Of the physical conditions for tJie visihility of Bodies \ 
and an adequate representation of their Forms. 

98. The most immediately obvious result of the exposure of 
a body to the light, is the existence, upon the body, of a line of 
shade separating the illuminated from the shaded portion of the 
body; and the production of a shadow upon any adjacent sur- 
face from which light is excluded by the given body. The con- 
struction of the lines of shade and of shadow have, accordingly, 
claimed attention in all the previous problems. 

But from (95) the curve of shade, alone, is insufficient, together 
with the apparent contour, to determine, unequivocally, the form 
of a body. This becomes further apparent, as follows: Having 
a given cylinder of rays, any curve traced upon its surface, may 
be the curve of contact of an infinite number of different bodies, 
all of which shall have the same projecting cylinder, and all of 
which will have this one assumed curve for their curve of shade. 
All these bodies, being inscribed tangentically in the same cylin- 
der of rays, will, moreover, cast identical shadows on any other 
given surface. 

99. See, at this point, PL XV., Fig. 55. KBEC is any opaque 
body. ABKD is a tangent cylinder of luminous rays, giving 
the curve of shade BHDG. Next, IFCJ represents a tangent 
projecting cylinder, or cylinder of visual rays reflected from the 
object, and gives the apparent contour FGrCH. Hence the por- 
tion, E — FGCH, of the body, bounded by that contour, and the 

8 



114 BRILLIANT POINTS. 

portion, GDH, of its curve of shade, are visible. But besides 
the curve and that contour, there is only our imagination to sug- 
gest the real configuration of the visible portions of the surface 
on which no contour lines are shown. These portions may 
indeed have any sinuous configuration, though experience and 
association lead us to assume that the figure represents an ellip- 
soidal body, or something like one. 

100. We therefore continue the examination of the action of 
light upon surfaces, in order to discover how their forms may be 
distinguished, and we find that a surface is illuminated in pro- 
portion to the directness with which the light falls upon it. See 
PL XY., Fig. 56. Here, let AB be the trace of a plane, tangent 
at T, to any curved surface. The space ah is illuminated by the 
beam of rays, M. The equal space cd, struck more obliquely 
by the light, is illuminated by the thinner beam N, while 
another equal space, ef. which is struck perpendicularly, receives 
all the light of the thickest beam, 0. Hence, in this case, the 
curved surface is most highly illuminated at the point of tan- 
gency, T. Along the curve of shade, tangent planes coincide 
with the direction of the light, and therefore the given surface 
there receives no light. The curve of shade therefore is the 
darkest part of the surface. 

101. Between the curve of shade and the brightest point, 
curves may be conceived lying on the surface, at all points of 
any one of which, tangent planes will make equal angles with the 
rays of light. 

But these angles represent the angles made by the surface 
itself with the light along the supposed curve. Therefore the 
curves just supposed will be curves of equal illumination, and 
the illumination of the surface will vary in intensity from the 
maximum darkness at the curve of shade, where the surface 
makes an angle of 0° with the light, to the maximum brightness, 
where the surface is normal to the light. 

Moreover, drawing aS, for example, the perpendicular width 
of the beam of light, it represents the sine of the angle Sha, made 
by the light with the tangent plane AB. Hence, in this view, 
the intensity with which a surface is illuminated is directly as the 
sine of the angle made by the light with that surface. 

102. But again : Bodies are not seen directly by the light 
thrown upon them, but by such portion of that light as is 
returned from them. 



SHADES AND SHADOWS. 115 

Hence, m general terms, the comparative amounts of light re- 
mitted to the eye, from the different points of a body, are, with 
its apparent contour, the means of judging of its form. 

103. Now, optical researches show, that, when light falls upon 
a body, a part is extinguished, or destroyed (absorbed according 
to the material theory), a part is reflected, and a third portion is 
polarized, by refraction among the molecules of the surface. 

104. Strictly speaking, a body is visible, only as it shows its 
own proper color, which it does by means of the refracted rays 
which it remits. 

Mirror-like surfaces reflect the rays just as they are received, 
and present images of all objects, from which light proceeds to 
fall upon them. They thus indicate their presence by their 
effects, but are strictly invisible, except that, in practice, there 
are no absolutely perfect mirrors ; hence they are faintly visible 
by the few refracted rays coming from the molecules of their 
surfaces. 

1 05. The law of reflection is, that the incident and reflected 
rays, at the same point of a surface, make equal angles with that 
surface. 

Hence, as there will, taking the general case of a double-curved 
surface, be but one such point, for a given fixed position of the 
luminous point, and point of sight, a perfectly polished double- 
curved body, exposed to a single source of light, would have but 
a single visible point ; and that, the one at which a normal line, 
or a tangent line or plane to the surface, would make equal angles 
with the incident ray, and the ray reflected to the eye. 

The condition for the complete visibility of a perfectly polished 
body, therefore, is, that it shall be exposed to light from all 
sources. 

106. The majority of the objects which present themselves to 
our observation, as a stone or piece of wood, are dull, or partially 
polished bodies, that is to say, their exterior, without failing to 
present an appearance of continuity, is nevertheless, by their 
porosity, broken up into minute asperities and cavities, as 
shown, greatly magnified, in PL XV., Fig. 57. These irregulari- 
ties, though separately invisible, from their minuteness as com- 
pared with the size of the body, yet, under the action of tne light, 
produce an aggregate effect, which is appreciable, since they are 
of considerable magnitude, as compared with the extreme tenuity 
of the molecules of light. Each asperity, therefore, is considered 



116 BRILLIANT POINTS. 

as having a number of indefinitely small facets, one or more of 
which is in a position to return light to the eve. 

But the exterior facets, lying in the perfectly continuous ima- 
ginary surface of the body, are larger than interior ones, which 
are withdrawn from polishing agencies ; and are nearer each other, 
as seen in projection, in the vicinity of the brilliant point, L, than 
in the more directly viewed portion of the continuous ideal sur- 
face containing them; and, besides, the properly disposed interior 
facets may often, in obliquely viewed regions, be hidden bv 
asperities in front of them, so that the point on a dull body, 
which corresponds to the only visible point of a perfectly 
polished body, is the brightest point of that dull body. 

107. Taking into the account the secondary remissions to the 
eye, from facets which are illuminated by reflection from other 
facets, it appears that the whole illuminated part of a dull body 
is rendered visible by a single source of light. 

Its shade, however, can become visible only by sending to the 
eye the light received from a secondary source, as from the 
atmosphere, or from surrounding bodies. 

Hence, if PL XY., Fig. 55, represents a dull body, illuminated 
by a single source of light, all that part, E — FGrDH, of the illu- 
minated part, which is in the field of vision, will be actually 
visible. The completely unilluminated shade, GCDH, though 
in the range of vision, will be invisible, since it remits no rays to 
the eye. 

108. But the theory of minute reflecting facets alone, is insuf- 
ficient, in not accounting for the colors of bodies ; since light that 
is merely reflected, retains the color which it had when coming 
in incident rays. It has therefore been presumed, and is con- 
firmed by experiments, that bodies are visible, mainly in conse- 
quence of light remitted by them after polarization by refraction, 
due to vibrations in their superficial molecules. These vibrations 
are supposed to be in unison — so to speak — with those of light 
of the color presented by those molecules ; and they make those 
molecules act as self-luminous points while illuminated. 

109. Hence, besides the contours furnished by their projec- 
tions, the conditions for the full representation of the continuity 
and forms of bodies, are, that their surfaces, being dull or porous, 
shall be illuminated from a primary and strong, and a lesser or 
reflected li^ht ; that their numberless facets shall be so distributed 
as to reflect increasing quantities of light, in successive rings of 



SHADES AND SHADOWS. 117 

equal reflection, from the curve of shade to the brilliant point, 
analogous to the curves of equal illumination (101) and that the 
vibrations of their superficial molecules, under the action of light, 
shall cause them to act, while exposed to the light, as self-lumi- 
nous bodies. 

Finally : a true representation of these apparent varying 
intensities of light and shade, exhibits graphically to the eye the 
continuity and the consecutive changes of form of surfaces, hence, 
in connection with the apparent contours of bodies, it adequately 
represents them to the eye. 

§ 3. — Of Brilliant Points and Lines. 

110. Most conspicuous and important, next to the line of 
shade of a body, is its brilliant point, or the element of greatest 
apparent illumination in its illuminated part. As these bril- 
liant points can, moreover, usually be constructed without diffi- 
culty, they will now be more fully considered. 

In treating of brilliant points thus in detail, the relative posi- 
tions of the luminous point, and the point of sight, with respect 
to a given body, are first to be noted. Since either of these 
points may be either at a finite or an infinite distance from the 
given body, the four following combinations may arise : 

Distance of the 
Luminous point. Point of sight. 

Infinite. Finite. 

Finite. " 

Infinite. Infinite. 

Finite. " 

111. The two former cases are properly treated under the 
head of " scenographie projections" or natural perspective, since 
the point of sight is there supposed to be at a finite distance from 
the object viewed. 

The two latter cases may here be discussed, since, whatever 
the distance of the source of light, the point of sight is at an 
infinite distance from the object, which is characteristic of ortho- 
graphic projections. 

112. In speaking now of the position of the brilliant point 
upon any surface, the distinction between the real, and the 



l paint must be observed. The real brilliant 
paint on a. given anrfara?, is the point at which the tangent plane 

I zat surface is peipendknlar to the direction, of the lights for 
this point is in a position to receive the greatest amount of light 
per :iz :: 5 :::;. :e 

if T, PL XV.. F:z\ ■:: :e the point of contact of a 
-.-_ •:.-.;- _ _ t AB^ to which th^ light ia normal, 
T wul be die real brilliant point of the given surface. 

IIS. The apparent brilliant point of a surface, is that point at 

which a normal to the surface bisects the angle between the 

:zz: : zz zzi z~r rez^czzi _ z iz ::: =azir ;z.l: 10a I ae 

izz-i.^iz :"^.::: ;:.:::• ::: ::_'.- :ir -:_:i :: 5 ircess^rT ;-: 

ziz — : - rlzl :: :_f z_ = zzd fxz: :--.:i : - Izzg 

11- JZ :':ZZ llr :Z^: : ITTrZ 5 7. ; Z ■ I r5 :il _ Z_ Zlr 

luminous point and point of sight both at an infinite distance, 
the brilliant point expands into a brilliant line, whose location 

will be explained in detail, in connection with the problems of 

the next section. 



s^:t::z :_ 

z 3rilliarLt Points and. Lines. 

.: : :.".-.' z - J 1 '"..- z 

v z ': j- • z"-l zz. * ~ zill :e z: zz'.t 



ii-zz~ z_ tt _tZ-tZ-_z -t t_-. ._ tne ci Tire rent 

depths of atmosphere, through which its different parts would 

- z^: ~T:i z zt~t t - rv-rrj part of it wonld appear 

Z _ : e : t : .: z ~ : : ". 1 : z z — : : : z: ::: : z z: r~ : : Z: : : .; 5 z : z z : z 
of soriaces, explained in connection with PL X~ 7:z : ". for 
:z^z :■. izz:crzzl zi: zzrzz:e :: :izz:z: zivze ~ ::.i _ :r5-z: 
a miiCw w diatiibatioz : : =■: iisposed as to reflect z - 5 

Z: t - ^ 

A plane, iHoininated bj divergin g rays -vill have a brilliant 



SHADES AND SHADOWS. 119 



Problem XLII. 

Tc find the brilliant point of a plane which receives light from 

a near luminous point. 

PI. XY, Fig. 58. Let the plane be vertical, and let TP be 
its horizontal trace ; and let S be the luminous point. Also let 
TR be the direction of the reflected rays, which are parallel (111). 

At any point, T, construct a line, TN, perpendicular to the 
given plane, and make the angle KTN=RTN. Then, through 
the luminous point, S, draw a ray, SB, parallel to KT, and B, 
its intersection with the plane TP, will evidently be the point at 
which the incident and reflected rays make equal angles with the 
normal to the plane. Hence B is the brilliant point required. 

Remarks. — a. Observe, that as NT is normal to the given 
plane, the plane of the incident and reflected rays is perpendicu- 
lar to the given plane. Hence, in this case, the ray SB is hori- 
zontal in space. 

b. The construction of the vertical projection of the figure, 
and the solution of the problem when the given plane is oblique 
to the planes of projection, may now be left to the student. 

§ 2. — Brilliant elements on Developable Surfaces. 

Passing to single curved surfaces, we shall first consider deve- 
lopable surfaces — and these, at first, as illuminated by parallel 
rays. 

Problem XLIII. 

To find the brilliant element on a cylinder, illuminated by 

parallel rays. 

First Solution.— PI. XY., Fig. 59. Let ANB— A'B' be a 
vertical right cylinder, and RO — R/O' the projections of a ray 
of light. OE represents the direction of those reflected rays 
which reach the eye. Now revolve the plane, R'O'E, of the 
incident and reflected rays, RO — R'O' and O r — OE, about its 
horizontal trace, O'E, into the horizontal plane of projection. 
The incident ray will then appear at R /7 ; — R /7/ 0, and L'"0 is 



J 20 BKILLIANT POINTS. 

the revolved position of the bisecting line of the angle EOE 
(105). Then, by making the counter revolution, this bisecting 
line will appear at LO — I/O', since it is in a plane B/O'E which 
is perpendicular to the vertical plane of projection. 

Now we may suppose that a row of asperities is ranged along 
the element 1ST — WW in which the bisecting line pierces the 
cylinder. If then each of these asperities has one or more 
minute facets which are perpendicular to LO — I/O', they will 
collectively reflect the rays contained in a vertical plane of rays 
through EO — K'O', and thus form a brilliant line N" — WW. 

Second Solution. — In this solution, some of the facets of the 
little asperities are supposed to be arranged in parallel elliptical 
bands, indefinitely narrow, and contained in the successive paral- 
lel planes of incident and reflected rays. 

See PL XV., Fig. 60, where the light is taken in the same 
direction as in the last figure, for the sake of easy comparison 
of the two. Those features of the construction, which are the 
same in both figures, are here omitted. 

KO is the revolved position of an incident ray, OE is a 
reflected ray, and DO is the bisecting line of their included angle. 
A ; B' is the vertical trace of a plane of rays, perpendicular to the 
vertical plane of projection, and determining the narrow elliptical 
band A'B', along which facets are supposed to be disposed so as 
to reflect light to the eye. A^CB^S is the revolved position of 
this elliptical band. On, the bisecting line of parallel chords, as 
Oe and db, which are perpendicular to DO, determines W", the 
point at which a normal, parallel to DO, can be drawn. 

In the counter revolution, W" returns to N"; hence, if the con- 
vex surface of the cylinder be supposed to be formed of consecu- 
tive bands of reflecting facets parallel to A'B, a vertical row of 
them will b3 found on an element N — N'N". 

Remarks. — a. It should be remembered, that the plane of the 
incident and reflected rays is normal to the reflecting surface. 

Otherwise, a ray striking a plane, at a point A, for example, 
might be reflected in a cone of reflected rays, generated by the 
revolution of the incident ray about a perpendicular to the plane 
at the point A. 

But in the case of an absolutely nnporous and polished cylin- 
der, or cone, which is struck obliquely by the light, a plane of 
incident and reflected rays, situated as in the present .example, 
cannot be made normal to the convex surface at N. Hence such a 



SHADES AND SHADOWS. 121 

cylinder, when illuminated from a single source of light, should 
be absolutely invisible throughout. 

The existence of asperities of surface, presenting reflecting 
facets, is, therefore, a necessary condition for the visibility of the 
surface thus illuminated. Hence reasoning or experiment must 
decide which of the theories of the location of the reflecting 
facets, given in the two solutions of this problem, is true, since 
these solutions give quite different locations to the brilliant line. 

The first construction is, I believe, the only one hitherto used, 
excepting the approximate one given in my "Elementary Pro- 
jection Drawing," and it agrees with the formation of the artifi- 
cial grain of the surface by the operation of turning. 

b. Either of the solutions of this problem may be applied, by 
the student, in finding the brilliant element of a cone. 



Pkoblem XLIY. 

To find the brilliant point of a cylinder which is illuminated 

by diverging rays, 

PI. XV., Fig. 61. Let L be the luminous point, and BDK the 
horizontal projection of the cylinder. The reflected rays being 
parallel, because the eye is at an infinite distance from the 
cylinder, let LGr be a line in the direction of the reflected rays. 

All lines through 0, and in a plane perpendicular to the axis 
of the cylinder, will be normal to its surface. The cylinder being 
vertical, in the present case, this plane will be horizontal, and 
EO, FO, GO, etc., are normals to the cylinder. 

Now, making L$=LE ; L&=LF, etc., the curve abc is the line 
at each of whose points incident rays, as L$, and reflected rays 
from $, etc., parallel to LGr, make equal angles with the normals, 
as E&O, etc., through the same points. 

Hence B, where this curve meets the convex surface of the 
cylinder, is the point of that surface at which the incident and 
reflected rays, LB and BY, make equal angles with the normal 
BO. That is, B is the horizontal projection of the required bril- 
liant point. Its vertical projection is not shown, it being found 
by merely projecting B into the circle cut from the cylinder by a 
horizontal plane through the luminous point. 



122 BRILLIANT POINTS. 



Problem XLV. 

To find the "brilliant point on a cone which is illuminated 
from a near luminous point. 

PL XV., Fig. 62. In the last problem, the curve in which the 
normals pierced the cylinder, coincided, in horizontal projection, 
with the projection BDK of the cylinder, and was, therefore, not 
constructed. The cone, not having a vertical surface, this curve 
of intersection of the normals would appear as a distinct curve, 
which must be constructed. 

Let VDTH — V'D'H be a cone of revolution, having its axis 
vertical; and let S,A' be the luminous point. 

Draw SA5", in the direction of the reflected rays. At S, any 
line will make equal angles with S5", and some incident ray, 
hence S is a point of the trial curve, containing the intersection 
of normals with incident rays. A is another point of the same 
curve, since SA, regarded first as an incident, and then as a 
reflected ray, makes the same angle with a normal VA at right 
angles to it. 

To find other points of the trial curve SAA. At any point, as 
V' ', on the line SJ", draw the horizontal projection, 5"V, of a 
normal. Kevolve this normal to the position hV, parallel to the 
vertical plane of projection. N'5', perpendicular to the element 
Y'D' — taken as the revolved position of the element Vv — will 
then be its vertical projection, and A'N' is the vertical projection 
(not drawn) of its primitive position. Now revolve this normal 
and the line S5", about an axis perpendicular to the vertical plane 
at N', and into the horizontal plane N'A". It will then appear 
at A"W— BY, and S5" will appear at A"— S'"B. Now make 
S'"^" — not drawn — equal to S //7 B, and h" will be the revolved 
position of another point of the trial curve. In the counter- 
revolution, h" returns in a short arc, h"h, parallel to B5", and A, 
on the primitive projection, 6"V, of the normal, is another point 
of the curve S A A. 

Having now found one point of this curve, in the manner just 
described, others are easily found as follows ; Revolve assumed 
points, as p and s, to p" and k, and join p" and h with Y. Then 
arcs, with radii S"y and S'"&, will locate points on p"Y and 
IcY — not shown — at distances from S //7 , equal to &"-p' f and S'"k. 



SHADES AND SHADOWS. 123 

Then, by counter-revolution, as before, find c and y. The trial 
curve SAcA can now be drawn. 

It is necessary, in the next place, to construct the curve in 
which all normals, which intersect S5", pierce the conic surface. 
Take, for example, the normal Yb". After revolution into the 
meridian plane AY, parallel to the vertical plane, this normal 
appears at Yb — b'r"', normal to the revolved position, YT) ', of 
the element W, which is in the meridian plane through Yb" . 
Hence r"' \r" is the revolved, and r,/ the primitive position of 
the intersection of this normal with the conic surface. 

Other similar points being found in like manner, the curve 
uer — e'n'r' may be sketched as the locus of the intersections of all 
normals, through points of A' — S5", with the conic surface. This 
curve intersects SAA, the locus of the intersections of normals 
with incident rays, at n,n\ This point is therefore that point, on 
the conic surface, at which the incident and reflected rays, Sn 
and nE, make equal angles with the normal, nY, at that point. 
Note, however, that these angles are not equal in projection, since 
their plane is oblique. 

Remark. — We may construct the point in which the normal at 
any point will pierce the line Sb". Thus, let F be the revolved 
position of any point at which a normal is to be drawn. FK, 
perpendicular to Y'H', is the revolved position of this normal. 
The revolution of FK, about Y'K as an axis, will generate a 
cone, whose surface will be normal to that of the given cone ; 
then the intersection of this auxiliary cone with the line Sb" — A', 
will give the point, which, when joined with Y,K, will be the 
normal required. 



§ 3. — Brilliant Points on Warped Surfaces. 

Peoblem XLYL 

To find the brilliant point of a warped surface, when illuminated by 

parallel rays. 

The solution of this problem involves the construction of a 
normal, parallel to a given line, L, viz., to the bisecting line of 
the angle included by an incident and a reflected ray. 

But since these rays make equal angles with a tangent plane 



12- BRILUJLNT PODSTS. 

wliich is perpendicular to the supposed normal, the construction 
of the normal may be effected by the construction of a tangent 
plane, perpendicular to a given line. But, again, since all planes 
which are perpendicular to the same straight line, are parallel, 
the operation, just proposed, is equivalent to the construction of 
a tangent plane parallel to a given plane. 

To construct the tangent plane, dr twc lines, A and E. 

not parallel, but each perpendicular to the given line. L. A 
system of tangent lines, parallel to A, will constitute a cylinder, 
tangent to the warped surface. Another such system, parallel to 
B, will form another tangent cylinder. The intersection of the 
curves of contact of these cylinders will be the point of tangency, 
on the warped surface, of a tangent plane perpendicular to L, and 
therefore the point of intersection, I, of the required normal, 
parallel to L. Hence I will be the required brilliant point. 



To construct the brilliant point on a screw, when illuminated by 

parallel rays. 

?r::V::-r. -.:. :'i —■:■", ; r v:::l r.-r-ifr.ts >:•- -r:-.A siie :::ir n.iri- 
dian plane parallel to the vertical plane, the axis of the screw 
being vertical, so as to draw, tangent to them, a sufficient arc 
that meridian curve of the screw, which lies in this merid 
plane. Then, as in Prob. XLL, find the bisecting line, IN, of 
angle included between an incident and a reflected ray. Revolve 
the line, M, till parallel to the vertical plane of projection. Draw 
a line, parallel to this revolved position, and normal to the meri- 
dian curve, just named, at a point, n. This point, n, will be the 
revolved position of the brilliant point In counter-revolution, 
it will return in a helical arc, and its projections will appear on 
the projections of N. 

T .is :oL5rm:rloz. '.:::. :e ~::;.;m: :i: :t :ie stuie^t. 



SHADES AND SHADOWS. 125 

§ 4. — Brilliant Points on Double-Curved Surfaces. 

Pboblem XL VIII. 

To find the brilliant point on any double-curved surface, when 
illuminated by diverging rays. 

Let PI. XV., Fig. 63, represent any doable-curved surface, 
not of revolution. Let S be the luminous point, and SR' a line 
parallel to the reflected rays. Let n'Nn" be the curve in which 
all normals, as RW, RN, and H"n", intersect the given surface. 
Also, let N'NN" be the locus of intersections of these normals 
with incident rays from S. That is, SN'^SR 7 , SN=SR, and 
SN'^SR", etc., so that if the rays, SN', etc., were drawn, the 
angles at N', etc., made by the incident ray SN', etc., and normals 
R'N"', etc., would be equal to the angles at R', etc., made by the 
reflected ray SR' with the same normals. Then N", the intersec- 
tion of the two curves just described, is that point on the given 
surface, where the incident and reflected rays make equal angles 
with the normal R1ST. Hence 1ST is the brilliant point. 

The practical difficulty, in case of the class of surfaces con- 
sidered in this article, and also in case of warped surfaces, is, that, 
unless they are surfaces of revolution, there is no convenient 
direct construction of the required normals. Hence the remain- 
ing problems embrace only the construction of the brilliant points 
of double-curved surfaces of revolution, when illuminated by 
parallel rays. 

When such surfaces are exposed to diverging rays, their 
brilliant points are found as in Prob. XLIV. 

Pboblem XLIX. 

To find the brilliant point on a sphere, illuminated by parallel rays. 

PI. IX., Fig. 26. A vertical plane, through the centre of the 
sphere, is here regarded as the vertical plane of projection. OL 
is the vertical projection of a ray through the centre of the sphere. 
As no horizontal projection of the sphere is shown, Or may be 
assumed as the ray, after being revolved into the vertical plane 



126 BRILLIANT POINTS. 

of projection, about LK, a line of that plane, as an axis. Then, 
also, OP will represent the direction of the reflected rays. Hence 
N'O, the revolved position of the bisecting normal of the angle 
rOP, gives W as the revolved position of the brilliant point. 
By a counter-revolution about LK, its primitive position, K", is 
found. 

Problem L. 
To find the brilliant point on a piedouche. 

PL X., Fig. 31. Ai — SV is any incident ray intersecting the 
axis A — T'A'. Let S' — AW represent the direction of the 
reflected rays, as the piedouche is seen in vertical projection. 
Revolving the assumed ray around AW, as an axis, till it is 
parallel to the horizontal plane of projection, it appears at SV" — 
Ai". Then draw A^, the bisecting line of the angle i"AW, 
and draw an auxiliary line, i"W. In the counter-revolution, 
H' % i'" returns to ifi, the line i"W returns to iW, and a to p. 
whence it is projected to p', because the plane WAz is perpen- 
dicular to the vertical plane of projection. ISTow^A and j/S' 
are the projections of the bisecting line of the angle VAW. The 
brilliant point is the point of contact of a tangent plane which 
is perpendicular to this bisecting line. Then revolve Ap — S '_£>', 
till parallel to the vertical plane, taking A — T'A 7 for an axis, 
when it will appear at Ap" — S'_p'". Next, draw the normal jl, 
parallel to p"'S\ and I will be the revolved position of the bril- 
liant point. Construct the projections of the horizontal circle 
through this point, and V,V', its intersection with the meridian 
plane through the bisecting line pA — p'A r , is the primitive 
position of the required brilliant point. 

Remark. The student may now construct the brilliant point 
on any of the double curved surfaces of revolution shown in the 
preceding plates, or upon a warped hyperboloid of revolution. 



SHAJ)ES AND SHADOWS. 12' 



CHAPTER II. 

OF THE REPRESENTATION OF THE GRADATIONS 
OF LIGHT AND SHADE. 

116. In observing a shade, or a shadow, two things arrest 
the attention, its position, and its intensity. 

To find the position of shades and of shadows upon surfaces, 
was the object sought in "Book I." Now, we are to determine 
their varying intensities, as affected by the several circumstances 
presently to be enumerated. Furthermore, the varying apparent 
intensity of the illumination of those parts of surfaces which 
are exposed to the light, will be affected by the same circum- 
stances. Hence the gradations of the lights, and of the shades 
and shadows of surfaces, will be discussed together. 

117. The particulars affecting apparent intensities, and the 
gradations of lights and shades, and which will next be dis- 
cussed in detail, are arranged under the following heads : 

1°. — Effects due to the laws of illumination and vision. 

2°. — Those due to the supposed infinite distances of objects 
from the eye. 

3°. — Those due to secondary sources of light; including the 
air as an illuminating medium. 

4°. — Those due to the nature of different surfaces. 

5°. — Those due to the atmosphere as an absorbing medium. 

6°. — Those due to variations in the intensity of the light. 

7°. — Those due to the forms of bodies. 

8°. — Those due (on the drawing) to drawing materials and 
processes of manipulation. 

118. Effects due to the laws of illumination and vision. 

a. Law of illumination. The intensity of light, or of the 
degree of actual illumination of the same surface, at different 
distances from the source of light, varies inversely as the square 
of the distance from the luminous source. 



128 THEORY OF SHADING. 

b. Case where the light comes from an infinite distance. The 
principle just stated applies, sensibly, whenever the source of 
light is at a finite distance. But when the light proceeds from 
an infinite distance, terrestrial bodies are all at substantially the 
same distance from it, and when similarly situated with respect 
to it, may be considered as equally illuminated. 

c. Law of apparent brilliancy. From remark a it follows, 
that if a surface be illuminated in a given degree, its apparent 
brilliancy should be inversely as its distance from the eye, but 
its image being reduced in the same proportion, the retina re- 
ceives light at the same rate or amount per superficial unit of its 
surface. Hence, practically, the apparent brilliancy of a surface 
of given brightness, and in a given position, will be the same at 
whatever distance it is seen ; if we disregard atmospheric 
effects. 

d. Illustration. The last remark is easily illustrated. See 
PI. XV, Fig. 64. 

Let A be a visible point of an illuminated surface, and let PQ 
be the pupil of the eye. The point A remits to the eye the cone 
of rays APQ, whose base is the pupil PQ. These rays converge 
in the eye, forming, on the retina at R, the image of the point 
A. If now the object be removed to A 7 , double the distance 
AC, the base, PQ, remaining of constant size, the section of the 
new cone, at the distance equal to AC from its vertex, will evi- 
dently have an area equal to one-fourth of PQ. But sections at 
a constant distance from their vertices measure the relative angu- 
lar capacities of the two cones. Hence the eye at double the dis- 
tance AC receives one-fourth as much light from the point A as at 
the distance AC. 

But to apply this result practically, we must consider more 
than a solitary point. Two points will be sufficient. Then 
let B be a second point on the body AB. The axis, Br, of its 
cone of rays gives r as the focus of those rays on the retina, and 
Rr as the space on the retina, illuminated by rays from the space 
AB on the object. Now remove AB to double the distance AC 
viz. to A'B', and the point B' will be imaged at r', one-fourth as 
brightly as at r. But Rr' is half of Rr, and hence the area, of 
which Rr' is the diameter, is one-fourth as large as that whose 
width is Rr. While, therefore, A'B' remits to the eye one-fourth 
of the light that AB remits, that light is received by the area, 
Rr', of the retina, which is one-fourth of the area Rr. Hence the 



PI. XV. 




SHADES AND SHADOWS. 429 

rate of illumination of the retina, for a unit of surface, is the 
same in both cases. 

e. In discussing this nice point, on which learned authors 
differ, care must be taken not to confound aggregate amount, 
and consequent dazzling effects of brilliancy, with its rate or inten- 
sity. A square foot of sun-lightened snow is as intensely bright, 
as to its degree of illumination, as a whole snow-bank, but the 
total amount of light from it, and extent of retina affected by it, 
is less than in case of the whole snow drift. Hence the amouni 
of dazzling effect is less. So a star, as brilliant as the sun, can 
easily be gazed upon, though its sensibly equal intensity of bright- 
ness with the sun is shown by its twinkling, which is its daz- 
zling effect upon a mere point of the retina; an effect which has 
only to be repeated in the same degree on many points at once 
of the retina, in order to become blinding. 

f. Shading of the penumbra. When a penumbra (85) is occa- 
sioned, as it is, by a near radiant body of sensible size, it receives 
light from a larger and larger portion of the luminous body, as 
it approaches its own outer boundary. Hence its shaded repre- 
sentation would be lighter and lighter in the same direction, 
unless prevented by the form of the surface containing it, or by 
other special circumstances. 

119. Effects due to the infinite distance of objects from the eye. 

a. Objects at an infinite distance seen as points. At the outset, 
and as a preliminary remark, it should be noted that supernatu- 
rally acute powers of vision are required to perceive an object 
at all, at an infinite distance. This may be proved by a refer- 
ence to the laws of vision, as follows : 

Each molecule of the surface of an illuminated object, having 
facets so disposed as to diffuse light in all directions, remits to 
the eye a cone of rays whose base is the pupil of the eye (118(f). 
This cone of rays, in passing through the lenses of the eye, con- 
verges to a point in its axis, and on the retina, which is the 
image of the point on the object from which the rays proceeded. 
Each point of the object thus produces its image, and these 
image points, collectively, form the superficial image of the 
object. 

But now conceive the object viewed to be at an infinite dis- 
tance from the eye. Frusta, of finite length, of all the cones of 
rays remitted to the eye, and having the pupil for their common 
base, will sensibly coincide in a single surface, which will be 

9 



130 THEORY OF SHADING. 

sensibly cylindrical. These frusta of rays, having thus a com 
mon axis, converge to a single common point on the retina, which 
is the image of the object. Now in order to be sensible of this 
image, consisting of but a single point, the retina must be of 
extraordinary sensibility — unless the light be of extreme intensity 
—also, its nervous texture must possess absolute continuity, in 
order to be impressible, strictly, at every point. 

b. Why represented as having sensible magnitude.— In the two 
respects just named, then, the eye must possess extraordinary 
power, in order to perceive infinitely remote objects; but, since 
they should appear as points, why are they represented as of 
sensible magnitude ? For the same reason that applies to ordinary 
vision. A giant nine feet high, and ninety feet distant, appears 
of the same size, nominally, or in a merely geometrical sense, as 
a boy three feet high, and thirty feet distant. But, practically, 
we say that each appears as we realize that it appears, and the 
appearance that we realize, i. e., the one that seems real to us, is 
determined by our knowledge, derived from other sources, of the 
real sizes of objects. Knowledge, therefore, of the actual sizes 
of objects, may make us believe that an infinitely remote object, 
seen with the organs above described, does appear of sensible 
size, according to that knowledge ; rather than as a point, accord- 
ing to its minute image on the retina. 

c. Theory of exaggeration of effects in shaded projections. But 
again, eyes of such acute sensibility, would also indicate with 
corresponding vividness the modifications of light and shade due 
to the causes already stated (117), hence the usual practice of 
greatly exaggerating effects, is quite rational in shading objects 
when shown in projection. 

This exaggeration appears principally in two particulars : 
First, in the extremely vivid contrast, between the element of 
shade and the brilliant point, which does not appear in common 
experience, but which appears natural, and agreeably suggestive 
of the forms of objects when represented in shaded projections. 
Second, in the effect allowed in representing surfaces at different 
distances. This is shown, first, in the manner of shading single 
surfaces seen obliquely ; and, second, in the tinting of a series of 
parallel surfaces, at slightly increasing distances, with tints of 
increasing darkness. 

120. Effects due to secondary sources of light; including tht 
air as an illuyainating medium. 



SHADES AND SHADOWS. 131 

• 

a. Direction of the strongest secondary light. A surface will 
diffuse b>ht in quantities proportioned to the amount received. 
It will receive most when struck perpendicularly. Therefore 
the greatest intensity of light from the particles of air, and of 
surrounding surfaces, will be in a direction opposite to that of 
the primitive rays. 

b. Of the relative darkness of different portions of the shade on a 
given surface. From (a) it follows that the shade of a body will 
have its faint brilliant element, due to the secondary lights, just 
as the illuminated part has its own brilliant element, due to the 
primary light. Therefore the shade is shaded lighter, in reced- 
ing from the line of shade, which is the darkest line of the body 
in reference to both lights. 

c. Of the relative darkness of different parts of a shadow. Any 
point in a shadow receives diffused light from surrounding 
objects, except within a cone whose vertex is the point, and 
whose base is the curve of contact of the cone with the body 
casting the shadow. Therefore, generally, the further a point 
of shadow is from the object casting it, the more diffused light 
it receives, and the lighter it is. 

The general principle just before named, must, however, be 
applied by inspection to each particular case of shadows on 
plane, convex, and concave surfaces. In case of a vertical prism 
casting its shadow on the horizontal plane, the shadow would 
be darkest at its junction with the base of the prism ; a result 
which is confirmed by observation. 

d. Optical, and actual effects of near surrounding objects. When 
an illuminated surface is placed quite near a shade, it actually 
illuminates a small neighboring region of that shade to a notice- 
able extent, and that region should therefore receive a lighter 
tint, in a drawing showing both bodies. 

Besides this actual effect, there are optical effects, due merely to 
the mutual influence of contrasts, as when a shadow falls on an 
illuminated surface, and when two tints of widely different in- 
tensity are Drought together. Here, the light surface appears 
lighter, and the dark surface darker, by contrast, as when black 
velvet laid upon black cloth makes the latter look lighter. 
Likewise, complementary colors heighten the effect of each other, 
when brought together. Thus, purple and gold, or blue and 
orange, each look more brilliant when side by side, than when 
seen separately. Such contrasts are presented in the shaded 



132 THEORY OF SHADING. 

drawing of an assemblage of objects, as well as by the original 
objects, hence the shading need not be exaggerated in order to 
produce the proper effect. Thus, if a shadow on a brilliant 
surface be of nearly uniform intensity, as it would be on a 
plane perpendicular to the light, its edges would, as they should 
do, appear darker than the interior portions, by contrast with the 
high light around them. 

e. Of the relative darkness of a shade and a shadow. Some 
interesting principles are suggested by considering the relative 
darkness of a shade and of a shadow, under various circum- 
stances. For example, let a shadow fall on the surface of a 
sphere. Any point in this shadow, will receive scattering rays 
of reflected and refracted light from all illuminated bodies and 
particles, in the hemisphere of space bounded by a tangent plane 
at the given point, and exterior to the cone, whose vertex is the 
same point, and whose base is its curve of contact with the body 
casting the shadow. A point in the shade of the sphere, near to 
which no other surface receives a shadow, will be illuminated by 
reflections from the entire hemisphere of space, bounded by the 
tangent plane at that point. Hence the latter point would be 
somewhat lighter than the former ; as also from the fact of its 
receiving the strongest atmospheric reflection (a). 

f In the case of the line of shade, as compared with a shadow 
on the illuminated part of a body, the shadow is the darker of 
the two. For, as the atmospheric reflections are strongest in a 
direction opposite to the light, they are weakest or nothing in 
the same direction as the light. Hence shadows cast upon sur- 
faces in the vicinity of their brilliant points are darker than the 
lines of shade of those surfaces. The shadow of the abacus 
upon the cylinder, for example, is darkest at the brilliant ele- 
ment of the cylinder, and fades away to the element of shade, 
where it disappears. 

g. When the primary and secondary lights are of nearly equal 
intensity, as seen in the comparatively uniform diffusion of light 
in a cloudy day, the contrast between the brilliant point, and the 
curve of shade, and the shadows, would be diminished. The 
latter would appear lighter, and the former darker, than when 
exposed to a single strong light. 

121. Effects due to the nature of surfaces. 

a. The nature of surfaces may vary, by being dull or polished, 
or by being differently impressed, in their molecules, by differeni 



SHADES AND SHADOWS. 133 

colored rays. While no surfaces are of absolutely perfect polish, 
owing to their porosity, yet, as they approximate to such a 
polish, it may be possible to express that fact graphically by the 
manner of shading them. 

A perfectly polished body, we have seen, has a single, well de- 
fined brilliant point. Hence, other things remaining the same, 
the more jpolished the surface, the smaller may be the region of lighl 
shading containing the brilliant point ; and the more dull the surface, 
the broader may be the area of light shading — as if each large 
asperity on a dull body, which is pretty directly illuminated, 
had its own somewhat conspicuous brilliant point. 

b. As already stated (104) bodies are, strictly speaking, not 
seen by their reflected light, but by that which they refract Re- 
flected light merely furnishes images of the objects which send 
rays to the reflecting surface, just as a surface echoes sound 
without imparting to it any new quality of tone derived from 
the echoing surface. 

In accounting for the colors of bodies, an analogy is supposed 
to exist between them and musical instruments, or resonant 
volumes of air. Many persons may have observed that upon 
sounding various notes, in a clear and strong humming manner, 
in a closed closet or small room, some of them will, without 
extra effort, sound much louder than others, as if the whole 
body of air in the room were excited to musical vibrations in 
unison with the loud sounding note. So surfaces, when exposed 
to the pulsations of light, seem to respond, in the vibrations of 
their molecules, only to the vibrations of rays of a certain color. 
By thus responding, these surfaces become, for the time, lumi- 
nous bodies, emitting, however, only such colored rays as their 
nature causes them to be excited by. 

Now, as bodies, in proportion to the perfection of their polish, 
reflect more and more light, less will remain to express their 
color by the refracting process just explained. Hence the higher 
the polish, the less clearly will the proper color of a surface be 
revealed. 

c. It may be added, as a matter of interest, to the illustration 
of the closed room, that such inclosed spaces are resonant, 
though in a less degree, to notes which differ but little from the 
principal note. So it is found, also, that the colors of bodies are 
not absolutely pure, but are always mixed with small portions 
of the neighboring colors of the spectrum. Thus the light from 



134 THEORY OF SHADING. 

a clear yellow flower, when dispersed by a prism, shows some 
orange and green rays, mixed with the pure yellow ones. 
122. Effects due to the atmosphere as an absorbing medium. 

a. Appearance of an illuminated surface seen obliquely. The 
atmosphere, not being perfectly transparent, extinguishes a por- 
tion of the light coming through it, and this portion is greater in 
proportion to the extent of air traversed. But we attribute to 
a surface the amount of light entering the eye in the direction 
from it to the eye. Hence the most remote portions of an illu- 
minated surface are the darkest in appearance. 

b. Appearance of a surface of shade seen obliquely. A surface of 
shade only remits to the eye tertiary rays, as they may be called, 
that is, ra)<s received by reflection from surrounding objects. 
Hence the atmospheric reflections coming to the eye in the direc- 
tion of the shaded object, may be supposed to be stronger than 
the light from the shaded object itself. But, as before, we attri- 
bute to the object the light coming to the eye in the direction of 
it, while the more remote it is, the greater is the body of inter- 
vening air which sends light from its own molecules to the eye. 
Hence the more remote a surface of shade is, the lighter it 
appears, or the nearer it appears to be of the same brilliancy of 
the atmosphere generally. This is very evident in nature, when 
we compare the lights and shades of distant buildings, or hills, 
with those of near ones. 

c. Results of the two preceding principles. 1°. A series of illu- 
minated parallel surfaces, seen perpendicularly, will appear 
darker and darker, as they become more distant. 2°. The 
reverse will happen, if these surfaces are in shade. 3°. Near 
shadows will appear intense, and remote ones faint. 4°. Portions 
of illuminated curved surfaces, which are more distant than the 
brilliant point, or line, will appear darker than that point or line. 
5°. Portions of the shade of curved surfaces, which are more 
distant than the line of shade, will appear lighter than the line 
of shade. 6°. Curved surfaces will appear of a more uniform 
tint, the more distant they are, the lights being darkened, accord- 
ing to (a), and the shades dimmed, according to (b). 

d. Limitation of the last principle. The last principle applies, 
however, to the relative depths of a body of air of finite thickness, 
through which bodies are seen from an infinite distance. See 
(119). 

e. Effect of the air on apparent zolor. The blue color of the air. 



SHADES AND SHADOWS. 135 

due to the rajs remitted from it by refraction, as explained already, 
is attributed to objects seen through it, as well as the intensity ot 
the atmospheric rays. Hence distant objects appear of a bluish 
cast, which increases in clearness with their distance. It is thus 
that, in a colored drawing, a distant object may be distinguished 
from the dull shading due to uniformly diffused light (120g). 

123. Effects due to variations in the intensity of the light. Inte- 
resting results follow an examination of the effect of lights, each 
received mainly from one direction, but varying in intensity. 
All so-called transparent media absorb some light, and, for the 
present purpose, it is sufficiently correct to assume that, for light 
passing through any given medium, and in greater quantity than 
can be absorbed by the medium, the absolute amount absorbed 
will be equal for all degrees of intensity. It is important here 
to observe that this absorbed light is, as it were, latent, and does 
not at all render the absorbing body visible. 

It follows that in case of a feeble light, where the total amount 
absorbed by the air and by surrounding objects is large in com- 
parison to the whole amount of light emitted, the diffused light 
caused by reflections and refractions among the particles of air, 
and the light remitted from surrounding objects, will both be 
small, but very little light will be thrown upon the shade, or 
into the indefinite shadow in space of a body, under these cir- 
cumstances, and hence this shade and shadow will approach in 
darkness to those formed under the supposition that they received 
no light, and were absolutely black (105). 

Moreover, the diffused light spoken of,, partially illumines 
objects within the indefinite shadow ; but if it be feeble, it will 
all be absorbed by those objects, and they will therefore, as 
shown above, not throw any light into the indefinite shadow, or 
upon the shade of the body. For this additional reason, there- 
fore, will the shades and shadows due to a feeble light be 
intensely dark, as compared with the illumined part of the body. 
This conclusion corresponds to the great blackness of the shades 
and shadows observable on and about buildings by moonlight. 
To return, now, to the case of clear sunlight, and then to apply 
these principles, it appears that, while the absolute amount of 
absorption remains the same, there is a large excess of light, 
available for producing diffused and reflected light, of sufficient 
brightness to produce reflection from bodies in the shadow upon 
the shade of an opaque body, so that enough light will reach 



136 THEORY OF SHADING. 

this shadow and shade, to mitigate their blackness considera 
bly. 

124. Effects due to the forms of bodies. 

a. Form of the curves of equal shade. The lines of equal tinl 
on different bodies, will be either similar or dissimilar. They 
will be similar on developable surfaces, where they are simply 
rectilinear elements, and on spheres, ellipsoids, etc., where they 
will generally be circles, ellipses, etc. But they will be dissimi- 
lar on the more complex surfaces, such as the torus and pie- 
douche. In the latter case, they will be gradually transformed, 
from a ring surrounding the brilliant point, to curves more and 
more nearly similar to the curve of shade. This should be 
remembered in distributing the tints upon such surfaces. 

b. Distribution of points of equal shade. Points of equal ap- 
parent brilliancy are not, however, symmetrically placed around 
the brilliant point as a centre. That is. on a sphere, for example, 
the curves of equal illumination are not circles, having their 
centres in the diameter through the brilliant point. See PI. 
XV., Fig. 65, which represents a sphere, with the plane of the 
paper taken as a plane of incident and reflected rays, as LO and 
OR'. Then B, the middle point of the arc AC, is the brilliant 
point. 

Now from (120a) the small superficial element, shown as 
having A for its centre, is the element having the greatest actual 
illumination. This element returns the broad beam of ravs, 
LA, partly by reflection (106) in the condensed beam AR. An 
equal element, equidistant from the brilliant point, as at C, re- 
ceives only the narrow beam of rays I/C, which, furthermore, 
it remits in the diffuse beam CR'. It is now quite evident, from 
a comparison of the breadth of the two reflected beams, that, if 
they diffused equal amounts of received light, the intensity of 
this diffused light, and consequently the apparent brilliancy of 
the elements A and C, would be inversely as the sine of the angle 
made by the reflected ray zuith a tangent plane at the point from which 
it proceeds. 

But not only is the beam CR' more diluted, so to speak, than 
AR, so far as the space filled by a given amount of light is con- 
cerned, but it actually contains fewer rays, viz., those of the 
narrow incident beam I/C. Hence, much more, is the element 
at A, brighter than the element C, which is at the same distance 
from the brilliant point, B. 



SHADES AND SHADOWS. 137 

In fact, the superior brilliancy of B over all other points, can 
be fully accounted for, in this view, only by taking account of 
the effect of distance (122a), and by supposing that some of the 
reflecting facets of molecules in the element A, which is seen 
obliquely, are hidden by others in advance of them, so that all 
the rays of the beam LA are not visibly reflected. 

Therefore, finally, from the above, we conclude, that on cylin- 
ders, cones, spheres, etc., points which are equidistant from the 
brilliant line, or point, should not receive equal tints ; and ob- 
serving, in Fig. 63, that tangent rays parallel to LA determine 
the curve of shade, that the darkest of two such points should 
be the one which is between the brilliant element and the visible 
projection of the element, or curve, of shade. 

c. Relative illumination of apparent contours. It may be sepa- 
rately noted that, as a sort of balancing between the results of 
the light received by a surface, the light reflected and refracted 
from it, and the obstruction of some of these remissions, there 
will be a narrow space along the apparent contour of curved 
surfaces, which will appear lighter than the portions a trifle 
further from that contour; for this space, which is very narrow 
in projection, is wide in reality, and so receives a broad beam 
of light ; which, however, it reflects within a narrow compass. 

d. Rate of variation of shades on different surfaces. A consi- 
deration of the forms of bodies shows that there must be varia- 
tions in the rate of gradation from light to dark, on different 
surfaces. For example, see PI. XV., Figs. 59 and 60, equal 
vertical strips of shade on the cylindrical surface will not appear 
equal in vertical projection, and the shading, used to express the 
curvature of its convex surface, would therefore proceed rapidly 
from dark to medium shades, and then, more gradually, from 
medium to light shades. But on the vertical plane seen oblique- 
ly, Fig. 58, similar equal strips would appear equal in projection, 
and the shading used to express the different distances of these 
strips from the eye (122), would proceed at a perfectly uniform 
rate from dark to light. 

e. The proper shading of hounding edges. The theory of the 
form of bounding edges, should be observed in the execution of 
shading. These edges are supposed not to be mathematical 
lines, but to be rounded, through the ultimate imperfections of 
workmanship, and the attrition of flying particles. Thus, an 
edge of a prism is to be regarded as minutely cylindrical ; and 



138 THEOEY OF SHADING. 

the circumference of the base of a cylinder is a portion of a 
torus, whose meridian section is a very minute quadrant. 

These edges, even when in shade, will have their own brilliant 
points, due to the scattering light which they may receive from 
surrounding objects, and to their superior smoothness, consequent 
on greater exposure to polishing agencies. Accordingly, very 
pleasing effects are produced, especially in an assemblage of 
shaded figures, as in a machine drawing, by shading the edges 
of each form, according to the principles here stated. 

f. Illustrations. First. On a shaded drawing of a vertical 
cylinder of revolution, the upper edge should everywhere be 
lighter than the bordering portions of the convex surface. The 
lower edge of its illuminated portion would be darker than the 
convex surface, but the lower edge of its shade should be lighter 
than the contiguous shade, though rather darker than the upper 
edge of the shade. 

Second. The successive parallel plane surfaces, forming the 
front of an edifice, are distinctly shown in projection, by thus 
treating their edges, which are mainly horizontal and vertical 
quarter cylinders. Upper and left hand edges, when exposed to 
the light, are left blank, or of a light shade, for a very narrow 
space. Lower and right hand edges, are quarter cylinders con- 
taining elements of shade, and may be ruled with a tint a little 
darker than that of the flat surface bounded by them. 

The light edges may be ruled with white, if not left blank in 
tinting. This course would produce the most striking effect in 
drawings on tinted paper. By thus distinguishing parallel sur- 
faces by the treatment of their edges, we avoid the necessity of 
an undue exaggeration of the difference between their shades, as 
explained in (122 c). 

125. Effects (on the drawing) due to drawing materials and 
processes of manipulation. 

a. Since the amount of remitted light received by the eye, 
both white, or reflected, and colored, or refracted, decreases in 
receding from the brilliant point, while the number of colored 
rays determines the apparent clearness and strength of color, it 
would seem that the projection of a body would be more accu- 
rately shaded by a mixture of India ink with its conventional tint, 
than by shading it with ink alone, to express its gradations of 
shades, and then covering it with a flat tint of the conventional 
color. For the latter method would give so little color compara 



SHADES AND SHADOWS. 139 

tively to the very dark shades, that it would be imperceptible, 
while in fact every part of the body does reveal its proper color 
to some extent. Otherwise, if the conventional flat tint were 
made strong enough to show, at the line of shade, it would obli- 
terate the light ink-shading beneath it, in the vicinity of the bril- 
liant point. In geometrical drawing, however, one of the exag- 
gerations, which compensate for the artificial character of all 
projections, is secured by the second method of treating colored 
objects. Hence the view presented in the " Elementary Projec- 
tion Drawing" (182). 

b. There are three simple methods which may be used, sepa- 
rately or together, in executing a shaded drawing. These 
methods are : 1. By softened tints. 2. By superposed flat tints. 
3. By touches. 

c. Shading in softened tints. This is difficult to execute, but 
would give, when perfectly done, the most perfectly smooth and 
continuous gradation of shade. Moreover, it requires to be done 
at one operation, with a mixture of ink and the appropriate con- 
ventional tint, as explained in (a). For if the two shades be 
made separately, though both by this same method of softened 
tints, discrepancies between their gradations will be apt to spoil 
the intended effect of each. 

d. Shading by superposed fiat tints. This consists in placing 
a narrow dark tint on the line of shade ; then, when dry, an- 
other, covering the first, and extending a little further, and so 
on. In this method, the drier the brush, the less will the edges 
of the successive bands show. 

In the last method, we rely somewhat on a free conformity 
of the softened shades to their theoretically true distribution, as 
determined by observation merely. 

In this method, we aim to have each of the well marked suc- 
cessive bands of tint exactly located so as to represent the bands 
of equal apparent brilliancy on the body represented. They 
should therefore, on account of their conspicuousness, be located 
according to precise knowledge of their proper position. 

But this location it is difficult, if not impossible, to effect, by 
any simple or generally available construction, owing to the 
complication of the problem, even in the simplest cases, as indi- 
cated in (124 b). 

e. Shading by touches. This method consists in placing 
directly upon each part of the drawing, a small quantity of ink, 



140 THEORY OF SHADING. 

with a tolerably dry brush, and of a darkness equal to that of 
the corresponding point of the surface represented. This method 
is quite similar to that of shading with a lead pencil, and, 
guided by practised judgment, will secure an orderly gradation 
of shades, without involving any of the raggedness and blotted 
appearance which are apt to appear in unskilful use of the two 
former methods. 

/. The first method is modified, by the size of the surface of 
paper to be shaded, into what may be called the methods by 
wet softened, tints, and by dry softened tints. The former is 
applicable to quite large surfaces, and requires a pretty large 
and wet brush. The latter is merely the second method with 
the edge of each tint softened by a brush damp with clear water. 

Neither the first nor the second method, alone, will ever pro- 
duce perfectly smooth shading. The third must be combined 
with them, so far as is necessary to fill up the irregularities of 
shade produced by those preliminary methods. The method by 
touches alone is very tedious, as applied to large surfaces. Hence 
it should then be combined with the second method. It is, 
however, conveniently applied, alone, to small surfaces. 

g. The great advantage of the method oy touches, is, that the 
minute light spots, finally left among the fine strokes of the 
brush, reproduce, in the drawing, the effect of the minute bril- 
liant points of the asperities of the object, and thus give the 
drawing an appearance of distinctness and solidity, which is 
very desirable. On the other hand, the uniform saturation of 
the paper with wet tints, used in the method of softened tints, 
produces a cloudy effect like that of the shade of polished 
bodies. Yet it does not, on the whole, adequately represent 
such bodies, because its gradation of shade is gradual, while 
theirs is apparently abrupt, on account of the dazzling bright- 
ness of their brilliant points. 

h. The effect of the minute brilliant points is also produced, 
in part, by shading on rough paper, on whose actual asperities 
the light may act. 

i. In the use of tinted paper, brilliant effects may be produced 
by using white tints for the brilliant points. 

Also, the effect of colored drawings will be heightened, by 
employing tinted paper of a color complementary to that whicfc 
prevails on the drawing. 



WARREN'S COMPLETE COURSE 



IN 



Descriptive Geometry, Stereotomy and Drawing, 

FOR SCHOOLS, COLLEGES, POLYTECHNIC INSTITUTES. 

Published by JOHN WILEY k SONS, 

53 East Tenth Street, New York. 



1. A PRIMARY GEOMETRY foe Common Schools. 

WITH SIMPLE AND PRACTICAL EXAMPLES IN PLANE AND PROJEC- 
TION DRAWING, AND SUITED TO ALL BEGINNERS. 

By S. Edward Warren, C.E., 

Author of a Series of Elementary and Higher Text-books on the Principles and 
Practice of Industrial Drawing. 

12mo, Cloth. 75 Cents. 

There is a growing and well-grounded feeling that there is an 
excess of attention to book and brain education, as compared with 
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In the present increasing reaction from the old abandoned 
apprenticeship system, this feeling manifests itself in a demand for 
increasing attention to Manual Training. 

Whether this manual training shall be in schools, or in real life, 
in shops, ships, railroads, offices or fields, Geometry, or the study 
of form, should, from the beginning, accompany arithmetic, or the 
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arrangement. 

Nearly all the Geometries hitherto published, and styled "Ele- 
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High School ; th,us depriving many thousands who need it, of an 
opportunity to study geometry. 



WARREN'S PRIMARY GEOMETRY, 

Prepared as the result of much experience and reflection, is, as 
indicated by its title, much more elementary than those commonly 
so called ; as may be gathered from the following statement of 

Its Distinctive Features. 

1. It presents the simplest geometrical truths, at first in a famil- 
iar conversational style, and is suited to the common schools in 
which so many, perhaps most, children end their school days. 

2. Its numerous and appropriate illustrations are original. 

3. It contains numerous suitable examples for practice. 

4. It embraces such a selection of truths of form, as well as 
measure, as to be especially adapted to accompany industrial draw- 
ing and other manual training. 

5. It is not encumbered and confused by embracing too little 
of each of too many and diverse applications. 



SPECIMENS OF MANY COMMENDATIONS. 

" Geometry might be taught much sooner than it is, and your book is well fitted 
to do it." — R. D. Dodge, Prin. Prospect Park Inst., Brooklyn, N. Y. 

" I am pleased with it as a practical work, and one that will make it natural and 
easy to introduce youth to the study."' — D. L. Kiehle, State Sup't Pub. List., Jdmn. 

" It impresses us very favorably. I fully agree with what you say in your pref- 
ace. — Richard Edwards, LL.D., State Sup't Pub. Inst., IU. 

" Prepared with great care. . . . More elementary than most Geometries. . . . 

Might be used in the grammar grade with good effect. . . ." — Xew England Journal of 

Education. » ^ 

"I like your Primary Geometry very much." — From a recent letter jrom a lady 

Teacher. 

PROFESSOR WARREN'S OTHER ELEMENTARY "WORKS, 

Suited to Academies and to Preparatory, High, Normal, Evening, 
and Industrial or Manual Schools are as follows : 

2. FREE-HAND GEOMETRICAL DRAWING, widely and 
variously useful in training the eye and hand in accurate sketching 
of plane and solid figures, lettering, geometric beauty, and design, 
etc. 12 folding plates, many cuts. Large l2mo, cloth, -$1.00. 



3. DRAFTING INSTRUMENTS AND OPERATIONS. A full 
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folding plates, many cuts. Large 12mo, cloth, $1.25. 

4. ELEMENTARY PROJECTION DRAWING. Fully explain- 
ing, in six divisions, the principles and practice of elementary 
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5. ELEMENTARY PERSPECTIVE. With numerous practical 
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vised edition. 2 plates. Large 12mo, cloth, $1.00. 

6. PLANE PROBLEMS on the Point, Straight Line, and Circle. 
225 problems. Many on Tangencies, and other useful or curious 
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mental constructions. For Colleges and Polytechnic Schools ; and 
for Architects, Engineers, Machinists, and Draftsmen. 150 wood* 
cuts, and plates. Large 12mo, cloth, $1.25. 



HIGHER WORKS. 

1. THE ELEMENTS OF DESCRIPTIVE GEOMETRY, SHAD- 
OWS AND PERSPECTIVE, with brief treatment of Trihedrals; 
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' ' Professor Warren here presents a most thorough and progressive course 
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Geometry, Shades and Per-; a condensed treatise, entirely new and 

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.5.% Pull descriptive catalogues and circulars, with references and testi- 
monials, on application. 



C 219 69 



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